1. Two Sum

1. Two Sum

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, 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].

解析

class Solution_1 {
public:
	// O(n^2)
	vector<int> twoSum(vector<int> &numbers, int target) {

		vector<int> vec;
		if (numbers.size()==0)
		{
			return vec;
		}

		for (int i = 0; i < numbers.size()-1;i++)
		{
			for (int j = i + 1; j < numbers.size();j++)
			{
				if (numbers[i]+numbers[j]==target)
				{
					vec.push_back(i);
					vec.push_back(j);
					break;
				}
			}
		}

		return vec;
	}

	// 使用一个哈希表(unorder_map)来解，第一遍扫描，保存到哈希表中，第二遍扫，看target-n在不在哈希表中，时间复杂度为O(n)。
        // 元素有重复的时候multi??
	vector<int> twoSum(vector<int> a, int target) {
		int i, j, k, l, m, n;
		map<int, int>mymap;
		map<int, int>::iterator it;
		vector<int>ans;
		for (i = 0; i < a.size(); i++){
			it = mymap.find(target - a[i]);
			if (it != mymap.end()){
				ans.push_back(it->second);
				ans.push_back(i);
				return ans;
			}
			else{
				mymap.insert(make_pair(a[i], i));
			}
		}
	}

	vector<int> twoSum1(vector<int>& nums, int target) {
		vector<int> vec;
		if (nums.size()==0)
		{
			return vec;
		}
		sort(nums.begin(), nums.end()); 
		int start = 0, end = nums.size()-1;
		while (start<end)
		{
			if (nums[start]+nums[end]==target)
			{
				vec.push_back(start);
				vec.push_back(end);
				break;
			}
			else if (nums[start] + nums[end] < target)
			{
				start++;
			}
			else
			{
				end--;
			}
		}
		return vec; //返回值的排序后的index,不符合题意
	}
};

题目来源

• leetcode 1. Two Sum
• 求和问题总结(leetcode 2Sum, 3Sum, 4Sum, K Sum)