Two Sum

[zhuangjinxin]

题目： Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

[zhuangjinxin]

解法1：

public int[] twoSum(int[] nums, int target) {
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[j] == target - nums[i]) {
                return new int[] { i, j };
            }
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

时间复杂度：O(n²) 空间复杂度：O(1)

[zhuangjinxin]

解法2：

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        map.put(nums[i], i);
    }
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement) && map.get(complement) != i) {
            return new int[] { i, map.get(complement) };
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

时间复杂度：O(n) 空间复杂度：O(n)

Hash把查找的时间复杂度从O(n)降低到O(1),但空间复杂度提高了；

[zhuangjinxin]

解法3：

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");
}

时间复杂度：O(n) 空间复杂度：O(n)

1.把解法2中两次循环合并； 2.map中只存在数组下标在i之前的元素，可以减少目标元素与i重复的判断；

[zhuangjinxin]

public static int[] twoSum(int[] nums, int target) {
    int[] result = new int[2];
    for (int i = 0; i <= nums.length-1; i++) {
        for (int j = i+1; j <= nums.length-1; j++) {
            if (nums[i]+nums[j]==target){
                result[0]=i;
                result[1]=j;
                return result;
            }
        }
    }
    return result;
}

对比了别人的写法，才知道自己的写法有多low～....

1. 很显然使用了最笨的两层循环的方法；
2. 先初始化返回数组result[],再赋值，再返回；(直接return new int[] {i,j}更简洁)
3. i<=nums.length-1等价于i<nums.length, 后者更简洁；
4. 未考虑到抛异常；