Leetcode-1. Two Sum-优快云博客

本文介绍了一个算法，用于在给定数组中找到两个数，它们相加等于特定的目标值。提供了两种实现方式：暴力遍历解法和使用哈希表优化的时间复杂度为O(n)的方法。

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].

AC代码：o(n^2)

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

AC代码：o(n) hash table

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> re;
        map<int,int> m;
        for(int i=0; i<nums.size(); i++)
            m.insert(map<int,int>::value_type(nums[i],i));
        for(int i=0; i<nums.size(); i++){
            int complement=target-nums[i];
            map<int,int>::iterator it=m.find(complement);
            if(it!=m.end() && it->second!=i){
                re.push_back(i);
                re.push_back(it->second);
                break;
            }
        }
        return re;
    }
};

总结：方法1为暴力遍历解法，时间复杂度高，方法二使用了map，空间复杂度提高，但时间复杂度为o(n);

map.find函数返回值为iterator类型，it->first = value,it->second = index.