【LeetCode】1. Two Sum 解题小结

 

 

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.

Example:

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

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

这题如果嵌套两个循环也可以做，不过时间复杂度是O(n²)。用unordered_map就要简单得多了。将元素值作为key，i作为value。如果在map中没有与遍历的当前元素之和为target，则把当前元素插入map中，这样至少会有一个元素在map中。不过需要注意返回的数组次序问题。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> inx(2);
        unordered_map<int, int> map;
        for (int i = 0; i < nums.size(); i++) {
            if (map.count(target - nums[i]))
            {
                inx[1]=i;
                inx[0]=map[target - nums[i]];
            }
            else
                map[nums[i]] = i;
        }
        return inx;
    }
};