leetcode_两数之和

https://leetcode-cn.com/problems/two-sum/solution/

1、简单暴力法

class Solution {
public:
    vector<int> twoSum(vector<int> &nums, int target);
};

vector<int> Solution::twoSum(vector<int> &nums, int target) {
    vector<int> ans;
    for(int i = 0; i < nums.size()-1; i++) {
        for(int j = i; j < nums.size(); j++) {
            if(nums[i] + nums[j] == target) {
                ans.push_back(i);
                ans.push_back(j);
                return ans;
            }
        }
    }
}

2、二次哈希法

class Solution {
public:
	vector<int> twoSum(vector<int> &nums, int target);
};

vector<int> Solution::twoSum(vector<int> &nums, int target) {
	vector<int> ans;
	map<int,int> hmap;
	for(int i = 0; i < nums.size(); i++) hmap[nums[i]] = i;
	for(int i = 0; i < nums.size(); i++) {
		int d = target - nums[i];
		if(hmap.find(d) == true && hmap[d]] != i) {
			ans.push_back(i);
			ans.push_back(hmap[d]);
			return ans;
		}
	}
}

遍历数组，

对每个元素进行考察时，　

因为target已知，

所以将问题转化为（target-正在考察的元素）是否在数组里。