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, 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].

 

思路：

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

 

代码：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int l = 0;
        int r = nums.size() - 1;
        int i = 0;
        vector<int> result;
        multimap<int, int> m;
        multimap<int, int>::iterator itmulti;
        for (vector<int>::iterator it = nums.begin(); it != nums.end(); it++) {
            int temp = *it;
            m.insert(make_pair(temp, i++));
        }
        sort(nums.begin(), nums.end());
        while (l < r) {
            if (nums[l] + nums[r] == target) {
                if (nums[l] == nums[r]) {
                    for (itmulti = m.equal_range(nums[l]).first;
                            itmulti != m.equal_range(nums[l]).second;
                            itmulti++) {
                        result.push_back((*itmulti).second);
                    }
                } else {
                    itmulti = m.equal_range(nums[l]).first;
                    result.push_back((*itmulti).second);
                    itmulti = m.equal_range(nums[r]).first;
                    result.push_back((*itmulti).second);
                }
                break;
            } else if (nums[l] + nums[r] < target) {
                l++;
            } else {
                r--;
            }
        }
        return result;
    }
};