【Day 57 】2022-12-27 - 1143.最长公共子序列

[azl397985856]

1143.最长公共子序列

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/longest-common-subsequence

前置知识

• 数组
• 动态规划

  题目描述

  给定两个字符串  text1 和  text2，返回这两个字符串的最长公共子序列的长度。

一个字符串的   子序列   是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。 例如，"ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。两个字符串的「公共子序列」是这两个字符串所共同拥有的子序列。

若这两个字符串没有公共子序列，则返回 0。

示例 1:

输入：text1 = "abcde", text2 = "ace" 输出：3
解释：最长公共子序列是 "ace"，它的长度为 3。 示例 2:

输入：text1 = "abc", text2 = "abc" 输出：3 解释：最长公共子序列是 "abc"，它的长度为 3。 示例 3:

输入：text1 = "abc", text2 = "def" 输出：0 解释：两个字符串没有公共子序列，返回 0。

提示:

1 <= text1.length <= 1000 1 <= text2.length <= 1000 输入的字符串只含有小写英文字符。

[Ryanbaiyansong]

··· class Solution: def longestCommonSubsequence(self, s: str, t: str) -> int:

最长公共子序列

    # 状态表示: f[i][j] 表示 s的前i个字符和t的前j个字符的公共子序列所有方案
    # 属性: max
    # 集合划分, 如果 s[i]和t[j] 相同的话
    # f[i][j] = f[i-1][j-1] + 1
    # 如果不同, 就是两者取max
    m, n = len(s), len(t)
    f = [[0] * (n+1) for _ in range(m+1)]

    for i in range(1, m+1):
        for j in range(1, n+1):
            if s[i-1] == t[j-1]:
                f[i][j] = f[i-1][j-1] + 1
            else:
                f[i][j] = max(f[i-1][j], f[i][j-1])

    return f[-1][-1]

···

[yuxiangdev]

class Solution { public int longestCommonSubsequence(String text1, String text2) { int m = text1.length(), n = text2.length(); int[][] dp = new int[m + 1][n + 1]; //初始化dp[0][] 和dp[][0]为0，因为当一个字符串长度为0时，那么他们的LCS长度也为0 Arrays.fill(dp[0], 0); for(int i = 0; i <= m; i++){ dp[i][0] = 0; }

    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j++){
            if(text1.charAt(i - 1) == text2.charAt(j - 1))//该字符可以加入LCS
                dp[i][j] = 1 + dp[i - 1][j - 1];
            else{//该位置的字符不相等，至少有一个不能加入LCS，先选择当前局部最优解，即选择前面的较大值
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }
    return dp[m][n];

}

}

[Ryanbaiyansong]

··· class Solution: def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int: n = len(status) q = deque() owned_keys = set() owned_boxes = set() for x in initialBoxes: if status[x]: q.append(x) else: owned_boxes.add(x) ans = 0 while q: t = q.popleft() ans += candies[t]

先处理钥匙, 如果钥匙能打开关闭的盒子的话, 将关闭的盒子加入队列中, 并且将此盒子移除出set

        # 如果不能, 将key先存下
        for x in keys[t]:
            if x in owned_boxes:
                q.append(x)
                owned_boxes.remove(x)
            else:
                owned_keys.add(x)

        for x in containedBoxes[t]:
            if status[x]:
                q.append(x)
            elif x in owned_keys:
                q.append(x)
                owned_keys.remove(x)
            else:
                owned_boxes.add(x)

    return ans

···

[JancerWu]

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length(), n = text2.length();
        int[][] f = new int[m+1][n+1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (text1.charAt(i-1) == text2.charAt(j-1)) {
                    f[i][j] = f[i-1][j-1] + 1;
                } else {
                    f[i][j] = Math.max(f[i-1][j], f[i][j-1]);
                }
            }
        }
        return f[m][n];

    }
}

[whoam-challenge]

class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: m, n = len(text1), len(text2) dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if text1[i - 1] == text2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + 1
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

    return dp[m][n]

[mayloveless]

思路

dp[i][j] 表示⻓度为i的t1子串和⻓度是j的t2子串的最⻓公共子序列⻓度 。递推公式：如果两个字母相等，则是上一个dp[i-1][j-1]+1，如果不相等，则取三种可能性最大的。

代码

/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function(text1, text2) {
    const n = text1.length
    const m = text2.length
    // dp[i][j]表示⻓度为i的t1子串和⻓度是j的t2子串的最⻓公共子序列⻓度 
    const dp = new Array(n+1).fill().map(() => new Array(m+1).fill(0));

    for (let i = 1;i<=n; i++) {
        for (let j = 1;j<=m;j++) {
            if (text1[i-1] === text2[j-1]) {
                dp[i][j] = dp[i-1][j-1] + 1;
            } else {
                // 3种可能
                dp[i][j] = Math.max(dp[i-1][j] , dp[i][j-1], dp[i-1][j-1]);
            }
        }
    }

    return dp[n][m];
};

复杂度

时间：O(nm) 空间：O(nm)

[yuexi001]

思路

动态规划

代码

C++ Code:

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.length(), n = text2.length();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; i++) {
            char c1 = text1.at(i - 1);
            for (int j = 1; j <= n; j++) {
                char c2 = text2.at(j - 1);
                if (c1 == c2) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
};

复杂度分析

• 时间复杂度：$O(mn)$
• 空间复杂度：$O(mn)$

[zjsuper]

class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int:

dp[i][j] means the length of common sequence till i in t1 and j in t2

    dp = [[0 for _ in range(len(text1)+1)] for _ in range(len(text2)+1)]
    ans= 0 
    for i in range(1,len(text2)+1):
        for j  in range(1,len(text1)+1):
            if text2[i-1] == text1[j-1]:
                dp[i][j] = dp[i-1][j-1] + 1
                ans = max(ans,dp[i][j])
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
    return ans 

[bookyue]

code

    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length();
        int n = text2.length();

        int[] dp = new int[n + 1];
        for (int j = 1; j <= m; j++) {
            int prev = dp[0];
            for (int i = 1; i <= n; i++) {
                int tmp = dp[i];
                if (text1.charAt(j - 1) == text2.charAt(i - 1))
                    dp[i] = prev + 1;
                else
                    dp[i] = Math.max(dp[i], dp[i - 1]);

                prev = tmp;
            }
        }

        return dp[n];
    }

[snmyj]

class Solution {
    public int longestCommonSubsequence(String t1, String t2) {
        char[] c1 = t1.toCharArray();
        char[] c2 = t2.toCharArray();
        int[][] dp = new int[c1.length + 1][c2.length + 1];
        for(int i = 1; i <= c1.length; ++i){
            for(int j = 1; j <= c2.length; ++j){

                if(c1[i - 1] == c2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }else{

                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[c1.length][c2.length];
    }
}

[testplm]

class Solution:
    def longestCommonSubsequence(self, s1, s2):
        dp = [[0 for i in range(len(s1)+1)] for j in range(len(s2)+1)]
        for i in range(1,len(dp)):
            for j in range(1,len(dp[i])):
                s1_sub = s1[:j]
                s2_sub = s2[:i]
                if s1_sub[-1] == s2_sub[-1]:
                    dp[i][j] = dp[i-1][j-1] + 1
                else:
                    dp[i][j] = max(dp[i][j-1],dp[i-1][j])
        return dp[-1][-1]

[Abby-xu]

思路

DP

代码

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2)
        dp, dpPrev = [0] * (n+1), [0] * (n+1)
        for i in range(1, m+1):
            for j in range(1, n+1):
                if text1[i-1] == text2[j-1]:
                    dp[j] = dpPrev[j-1] + 1
                else:
                    dp[j] = max(dp[j-1], dpPrev[j])
            dp, dpPrev = dpPrev, dp
        return dpPrev[n]

复杂度

...

[Elsa-zhang]

class Solution:
    def longestCommonSubsequence(self, A: str, B: str) -> int:
        m,n = len(A), len(B)
        ans = 0
        dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if A[i-1] == B[j-1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                    ans = max(ans, dp[i][j])
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        return ans

[FlipN9]

/**
    subsequence: 没有对字母顺序的要求
    TC: O(NM)   SC: O(NM)
*/

// SC 优化 成 O(M): 滚动数组
class Solution {
    public int longestCommonSubsequence(String s1, String s2) {
        int N = s1.length(), M = s2.length();
        int[][] dp = new int[2][M + 1];
        for (int i = 1; i <= N; i++) {
            int a = i & 1, b = (i - 1) & 1;
            char c1 = s1.charAt(i - 1);
            for (int j = 1; j <= M; j++) {
                if (c1 == s2.charAt(j - 1))
                    dp[a][j] = 1 + dp[b][j - 1];
                else
                    dp[a][j] = Math.max(dp[a][j - 1], dp[b][j]);
            }
        }
        return dp[N & 1][M];
    }
}

class Solution1 {
    public int longestCommonSubsequence(String s1, String s2) {
        int N = s1.length(), M = s2.length();
        // dp[N][M] = len of longest subseq from s1[0... N - 1] s2[0... M - 1]
        int[][] dp = new int[N + 1][M + 1];
        // Base case: dp[0][0] = 0;
        for (int i = 1; i <= N; i++) {
            for (int j = 1; j <= M; j++) {
                if (s1.charAt(i - 1) == s2.charAt(j - 1))
                    dp[i][j] = 1 + dp[i - 1][j - 1];
                else
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
            }
        }
        return dp[N][M];
    }
}

[chenming-cao]

解题思路

动态规划。定义二维数组dp，dp[i][j]表示从头开始的长度为i的text1的子字符串和从头开始长度为j的text2的子字符串的LCS的长度。状态转移方程：分为两种情况，如果对应index上的字符相同，那么dp[i][j] = dp[i - 1][j - 1] + 1，如果对应index上的字符不同，那么dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。用双层循环最后求得dp[text1.length()][text2.length()]即为最终结果。

代码

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int len1 = text1.length();
        int len2 = text2.length();
        int[][] dp = new int[len1 + 1][len2 + 1];
        // dp[i][j] corresponds to the length of LCS between the substring of text1 starting at index 0 with length i and the substring of text2 starting at index 0 with length j

        char[] arr1 = text1.toCharArray();
        char[] arr2 = text2.toCharArray();

        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                // if characters in text1 and text 2 are the same
                if (arr1[i - 1] == arr2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
                // if characters are different
                else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[len1][len2];
    }
}

复杂度分析

• 时间复杂度：O(m * n)，m为text1的长度，n为text2的长度
• 空间复杂度：O(m * n)

[tiandao043]

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int n1=text1.size(),n2=text2.size();
        vector<vector<int>> dp(n1+1,vector<int>(n2+1));
        for(int i=0;i<n1;i++){
            for(int j=0;j<n2;j++){
                if(text1[i]==text2[j]){
                    dp[i+1][j+1]=dp[i][j]+1;
                }else{
                    dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);
                }
            }
        }
        return dp[n1][n2];
    }
};

[GG925407590]

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.length(), n = text2.length();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; i++) {
            char c1 = text1.at(i - 1);
            for (int j = 1; j <= n; j++) {
                char c2 = text2.at(j - 1);
                if (c1 == c2) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
};

[klspta]

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length();
        int n = text2.length();
        int[][] dp = new int[m + 1][n + 1];
        int res = 0;
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
                if(text1.charAt(i - 1) == text2.charAt(j - 1)){
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 1] + 1);
                } 
            }
        }
        return dp[m][n];
    }
}

[saitoChen]

思路

dp[i][j]指的是字符串text1截止到i和字符串text2截止到j的最长公共子序列的长度 递推：当text1[i - 1] === text2[j - 1]时有 -> dp[i][j] = dp[i - 1][j - 1] + 1，当text1[i - 1] !== text2[j - 1]时有 -> dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1])

代码

const longestCommonSubsequence = (text1, text2) => {
    const len1 = text1.length, len2 = text2.length
    // 取len1+1,len2+1的原因是要甩出i - 1, j - 1的量
    const dp = new Array(len1 + 1).fill(0).map(() => new Array(len2 + 1).fill(0))
    for (let i = 1; i <= len1; i++) {
        for (let j = 1; j <= len2; j++) {
            if (text1[i] == text2[j]) {
                dp[i][j] = dp[i - 1][j - 1] + 1
            } else {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1])
            }
            console.log('i ->', i, 'j ->', j, 'dp->', dp[i][j])
        }
    }
    return dp[len1][len2]
}

复杂度分析: 时间复杂度：O(N^2) 空间复杂度：O(N)

[zywang0]

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length(), n = text2.length();
        int[][] f = new int[m+1][n+1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (text1.charAt(i-1) == text2.charAt(j-1)) {
                    f[i][j] = f[i-1][j-1] + 1;
                } else {
                    f[i][j] = Math.max(f[i-1][j], f[i][j-1]);
                }
            }
        }
        return f[m][n];

    }
}

[buer1121]

class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: n1=len(text1)+1 n2=len(text2)+1 dp=[[0 for in range(n2)] for in range(n1)]

    for i in range(1,n1):
        for j in range(1,n2):
            if text1[i-1]==text2[j-1]:
                dp[i][j]=dp[i-1][j-1]+1
            else:
                dp[i][j]=max(dp[i-1][j],dp[i][j-1])

    return dp[-1][-1]

[Moin-Jer]

class Solution { public int longestCommonSubsequence(String t1, String t2) { char[] c1 = t1.toCharArray(); char[] c2 = t2.toCharArray(); int[][] dp = new int[c1.length + 1][c2.length + 1]; for(int i = 1; i <= c1.length; ++i){ for(int j = 1; j <= c2.length; ++j){

            if(c1[i - 1] == c2[j - 1]){
                dp[i][j] = dp[i - 1][j - 1] + 1;
            }else{

                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }
    return dp[c1.length][c2.length];
}

}

[paopaohua]

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length();
        int n = text2.length();
        //初始化
        int[][] dp = new int[m + 1][n + 1];
        //遍历顺序
        for(int i = 1; i <= m;i++){
            for(int j = 1;j <= n;j++){
                if(text1.charAt(i - 1) == text2.charAt(j - 1)){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i - 1][j],dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
}

[RestlessBreeze]

代码

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.length(), n = text2.length();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; i++) {
            char c1 = text1.at(i - 1);
            for (int j = 1; j <= n; j++) {
                char c2 = text2.at(j - 1);
                if (c1 == c2) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
};

[Jetery]

1143. 最长公共子序列

思路

动态规划

代码 (cpp)

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.size(), n = text2.size();
        vector<vector<int>> dp(m + 1, vector(n + 1, 0));
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (text1[i - 1] == text2[j - 1]) 
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                else 
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[m][n];
    }
};

复杂度分析

• 时间复杂度：O(n * m)
• 空间复杂度：O(n * m)

[BruceZhang-utf-8]

代码

• 语言支持：Java

Java Code:

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        char[] t1 = text1.toCharArray();
        char[] t2 = text2.toCharArray();
        int length1 = t1.length;
        int length2 = t2.length;
        int[][] dp = new int[length1+1][length2+1];
        for (int i = 1; i < length1 +1; i++) {
            for (int j = 1; j < length2 +1; j++) {
                if (t1[i-1] == t2[j-1]){
                    // 找到一个 lcs 的元素，继续往前找
                    dp[i][j] = 1+ dp[i-1][j-1];
                }else {
                    dp[i][j] = Math.max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        return dp[length1][length2];
    }
}

[MaylingLin]

思路

---

dp建模套路

代码

class Solution:
    def longestCommonSubsequence(self, A: str, B: str) -> int:
        m, n = len(A), len(B)
        ans = 0
        dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if A[i - 1] == B[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                    ans = max(ans, dp[i][j])
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        return ans

复杂度

---

• Time: $O(m*n)$，m和n分别为A和B的长度。
• Space: $O(m*n)$

[AtaraxyAdong]

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int n = text1.length();
        int m = text2.length();
        char[] charArray1 = text1.toCharArray();
        char[] charArray2 = text2.toCharArray();
        int[][] dp = new int[n + 1][m + 1];
        int ans = 0;

        for (int i = 1; i < n + 1; i++) {
            for (int j = 1; j < m + 1; j++) {
                if (charArray1[i-1] == charArray2[j-1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    ans = Math.max(ans, dp[i][j]);
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return ans;
    }
}

[Ryanbaiyansong]

class Solution:
    def longestCommonSubsequence(self, s: str, t: str) -> int:
        m, n = len(s), len(t)
        f = [[0] * (n + 1) for _ in range(m + 1)]

        for i in range(1, m+1):
            for j in range(1, n+1):
                if s[i-1] == t[j-1]:
                    f[i][j] = f[i-1][j-1] + 1
                else:
                    f[i][j] = max(f[i-1][j], f[i][j-1])

        return f[-1][-1]

[chiehw]

思路：动态规划

在 text1 取长度为 i 的字符串，在 text2 取长度为 j 的字符串，并记录下他们的子串长度。每次改变一个字符，可以通过前面记录过的子串长度来推导。

• 如果最后一个字符相同：新的子串长度等于 lens[i-1][j-1] + 1。
• 如果最后一个字符不同：新的子串长度等于 max(lens[i-1][j], lens[i][j-1])

代码

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int len1 = text1.length(), len2 = text2.length();
        vector<vector<int>> lens(len1 + 1, vector<int>(len2 + 1));

        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if (text1[i - 1] == text2[j - 1]) {
                    lens[i][j] = lens[i - 1][j - 1] + 1;
                } else {
                    lens[i][j] = max(lens[i - 1][j], lens[i][j - 1]);
                }
            }
        }

        return lens[len1][len2];
    }
};