Open carloscn opened 1 year ago
先统计1的数量count_1, 0的数量就是 len - count_1 。这样分割之后就可以用数学运算的方式代替每一次的遍历了。
开始分割字符串:
遍历 左字符串找0的数量为l_count_0,1的数量就为 index + 1 - l_count_0, 那么由字符串剩余1的数量就是 count_1 - (index + 1 - l_count_0),然后求和即可。
fn count_0(v:&Vec<char>, index:usize) -> usize
{
let mut ret:usize = 0;
for i in 0..(index + 1) {
if v[i] == '0' {
ret += 1;
}
}
return ret;
}
pub fn max_score(s: String) -> i32
{
if s.is_empty() {
return 0;
}
let mut count_1:usize = 0;
let s_vec:Vec<char> = s.chars().collect();
for e in &s_vec {
if *e == '1' {
count_1 += 1;
}
}
let mut max_count:usize = 0;
for i in 0..s.len() {
let l_count_0:usize = count_0(&s_vec, i);
let r_count_1:usize = count_1 - (i + 1 - l_count_0);
if max_count < l_count_0 + r_count_1 {
max_count = l_count_0 + r_count_1;
}
}
return max_count as i32;
}
问题分析
给你一个由若干 0 和 1 组成的字符串 s ,请你计算并返回将该字符串分割成两个 非空 子字符串(即 左 子字符串和 右 子字符串)所能获得的最大得分。
「分割字符串的得分」为 左 子字符串中 0 的数量加上 右 子字符串中 1 的数量。
示例 1: 输入:s = "011101" 输出:5 解释: 将字符串 s 划分为两个非空子字符串的可行方案有: 左子字符串 = "0" 且 右子字符串 = "11101",得分 = 1 + 4 = 5 左子字符串 = "01" 且 右子字符串 = "1101",得分 = 1 + 3 = 4 左子字符串 = "011" 且 右子字符串 = "101",得分 = 1 + 2 = 3 左子字符串 = "0111" 且 右子字符串 = "01",得分 = 1 + 1 = 2 左子字符串 = "01110" 且 右子字符串 = "1",得分 = 2 + 1 = 3
示例 2:
输入:s = "00111" 输出:5 解释:当 左子字符串 = "00" 且 右子字符串 = "111" 时,我们得到最大得分 = 2 + 3 = 5
示例 3:
输入:s = "1111" 输出:3
提示:
2 <= s.length <= 500 字符串 s 仅由字符 '0' 和 '1' 组成。
来源:力扣(LeetCode) 链接:https://leetcode.cn/problems/maximum-score-after-splitting-a-string