#Leetcode# 1.Two Sum

https://leetcode.com/problems/two-sum/description/

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].

时间复杂度：$O(N)$

题解：使用 count ，返回的是被查找元素的个数，如果有，返回 1；否则返回 0；只遍历一个数字，另一个数字提前存起来

代码：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> mp;
        vector<int> res;
        int n = nums.size();
        for(int i = 0; i < nums.size(); i ++) {
            mp[nums[i]] = i;
        }
        
        for(int i = 0; i < nums.size(); i ++) {
            int t = target - nums[i];
            if(mp.count(t) && mp[t] != i) {
                res.push_back(i);
                res.push_back(mp[t]);
                break;
            }
        }
        return res;
    }
};

代码（JAVA）：

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] ans = new int[2];
        HashMap<Integer, Integer> mp = new HashMap<>();
        for(int i = 0; i < nums.length; i ++) {
            int t = nums[i];
            int pos = target - t;
            if(mp.containsKey(pos)) {
                ans[0] = Math.min(i, mp.get(pos));
                ans[1] = Math.max(i, mp.get(pos));
                break;
            }
            mp.put(nums[i], i);
        }
        return ans;
    }
}

　　

　

打开新世界的门缝？？？