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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]
思路：遍历数组，对每个元素 x，查找（target-x）是否存在于数组中。
代码如下：（参考了九章算法代码 http://www.jiuzhang.com/solutions/two-sum/）
1 class Solution { 2 public: 3 vector<int> twoSum(vector<int>& nums, int target) { 4 5 unordered_map<int,int> hash; 6 vector<int> result(2,-1); 7 8 for(int i=0; i<nums.size(); i++) 9 { 10 if(hash.find(target-nums[i]) != hash.end()) 11 { 12 result[0]=hash[target-nums[i]]; 13 result[1]=i; 14 return result; 15 } 16 17 hash[nums[i]]=i; 18 } 19 20 return result; 21 22 23 } 24 25 26 };
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原文地址：http://www.cnblogs.com/gexinmen/p/5931886.html
