LeetCode 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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

UPDATE (2016/2/13):

The return format had been changed to zero-based indices. Please read the above updated description carefully.

分析：

方法 1:暴力，复杂度 O(n2)，会超时

方法 2:hash。用一个哈希表，存储每个数对应的下标，复杂度 O(n).

方法 3:先排序，然后左右夹逼，排序 O(n log n)，左右夹逼 O(n)，最终 O(n log n)。但是注

意，这题需要返回的是下标，而不是数字本身，因此这个方法行不通。

代码：

//LeetCode, Two Sum
// 方法 2:hash。用一个哈希表，存储每个数对应的下标 // 时间复杂度 O(n)，空间复杂度 O(n)
class Solution {
public:
vector<int> twoSum(vector<int> &num, int target) {
unordered_map<int, int> mapping;
vector<int> result;
for (int i = 0; i < num.size(); i++) {
mapping[num[i]] = i;
}
for (int i = 0; i < num.size(); i++) {
const int gap = target - num[i];
if (mapping.find(gap) != mapping.end() && mapping[gap] > i) {
result.push_back(i + 1);
result.push_back(mapping[gap] + 1);
break;
} }
return result;
}
};