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Given an array of integers, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are NOTzero-based.
You may assume that each input would have exactly one solution
numbers=[2, 7, 11, 15], target=9
return [1, 2]
Analyse: Be aware of that one element could appear more than once. Map each element into an array.
Runtime: 10ms
1 class Solution { 2 public: 3 /* 4 * @param numbers : An array of Integer 5 * @param target : target = numbers[index1] + numbers[index2] 6 * @return : [index1+1, index2+1] (index1 < index2) 7 */ 8 vector<int> twoSum(vector<int> &nums, int target) { 9 // write your code here 10 11 // use a hash map to store array information 12 // the key is element in nums, value is the index of that element 13 unordered_map<int, vector<int> > um; 14 for (int i = 0; i < nums.size(); i++) { 15 um[nums[i]].push_back(i); 16 } 17 18 // scan the array and find if (target - element) in the hash map 19 vector<int> result; 20 for (int i = 0; i < nums.size(); i++) { 21 if (um.find(target - nums[i]) != um.end()) { 22 // a value appears more than once and happens to equal to half of target 23 if (nums[i] == target - nums[i]) { 24 result.push_back(um[nums[i]][0] + 1); 25 result.push_back(um[nums[i]][1] + 1); 26 } 27 else { 28 int index1 = um[nums[i]][0], index2 = um[target - nums[i]][0]; 29 result.push_back(min(index1, index2) + 1); 30 result.push_back(max(index1, index2) + 1); 31 } 32 break; 33 } 34 } 35 return result; 36 } 37 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/5797193.html
