标签：code air example tar ber begin imap int size
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].
1.使用multimap存储元素和对应下标
2.对数组进行排序
3.从数组两端寻找目标元素
4.从multimap中查找对应下标
1 class Solution { 2 public: 3 vector<int> twoSum(vector<int>& nums, int target) { 4 int l = 0; 5 int r = nums.size() - 1; 6 int i = 0; 7 vector<int> result; 8 multimap<int, int> m; 9 multimap<int, int>::iterator itmulti; 10 for (vector<int>::iterator it = nums.begin(); it != nums.end(); it++) { 11 int temp = *it; 12 m.insert(make_pair(temp, i++)); 13 } 14 sort(nums.begin(), nums.end()); 15 while (l < r) { 16 if (nums[l] + nums[r] == target) { 17 if (nums[l] == nums[r]) { 18 for (itmulti = m.equal_range(nums[l]).first; 19 itmulti != m.equal_range(nums[l]).second; 20 itmulti++) { 21 result.push_back((*itmulti).second); 22 } 23 } else { 24 itmulti = m.equal_range(nums[l]).first; 25 result.push_back((*itmulti).second); 26 itmulti = m.equal_range(nums[r]).first; 27 result.push_back((*itmulti).second); 28 } 29 break; 30 } else if (nums[l] + nums[r] < target) { 31 l++; 32 } else { 33 r--; 34 } 35 } 36 return result; 37 } 38 };
标签：code air example tar ber begin imap int size
原文地址：http://www.cnblogs.com/sindy/p/6413852.html
