标签：log pre widgets end air turn target back order
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].
解法一：
1 class Solution { 2 public: 3 vector<int> twoSum(vector<int>& nums, int target) { 4 unordered_map<int, int> u_map; 5 vector<int> result; 6 7 for (int i = 0; i < nums.size(); ++i) 8 { 9 if (u_map.find(target - nums[i]) != u_map.end()) 10 { 11 result.push_back(u_map[target - nums[i]]); 12 result.push_back(i); 13 return result; 14 } 15 16 u_map.insert(make_pair(nums[i], i)); 17 } 18 19 return result; 20 } 21 };
map搞一把
标签：log pre widgets end air turn target back order
原文地址：http://www.cnblogs.com/abc-begin/p/7536110.html
