LeetCode Two Sum（001）解法总结Given an array of integers, return

描述

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].

思路

先使用for循环确定当前数字，再嵌套一个for循环匹配第二个数字，一旦满足要求，就进行返回。

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] out = new int[2];
        for(int i =0;i<nums.length;i++){
            boolean flag = false;
            for(int j =i+1;j<nums.length;j++){
                if(nums[i] + nums[j] == target){
                    out[0] = i;
                    out[1] = j;
                    flag = true;
                    break;
                }
            }   
            if(flag){
                break;
            }
        }
        return out;
    }
}

可以暴力AC，但是时间复杂度和空间复杂度都不是很好。

Runtime: 59 ms, faster than 18.99% of Java online submissions for Two Sum.

Memory Usage: 39.3 MB, less than 6.43% of Java online submissions for Two Sum.

优化

看到问题讨论区有使用map作为辅助查询工具的，可以大大提高数据检索速度。

public int[] twoSum(int[] numbers, int target) {
    int[] result = new int[2];
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    for (int i = 0; i < numbers.length; i++) {
        if (map.containsKey(target - numbers[i])) {
            result[1] = i;
            result[0] = map.get(target - numbers[i]);
            return result;
        }
        map.put(numbers[i], i);
    }
    return result;
}

该算法的思路是：使用for循环进行遍历，当前数据为i所记录的位置，但是不同的是它进行前向查找，将当前数据和之前已经完成遍历的数据匹配。这样做的好处在于可以将已遍历过的数据加入map集合，方便进行比对、节约检索时间和回退时间，且不会出现二次匹配成功的问题。

Runtime: 1 ms, faster than 99.90% of Java online submissions for Two Sum.

Memory Usage: 42.4 MB, less than 5.65% of Java online submissions for Two Sum.

可以看到时间复杂度大幅提高，空间开支增加。