Two Sum

Question

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 not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

My Solution

#include <algorithm>

class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
/**
* 变量：
* mid:中值；pBig:<span style="font-family: Arial, Helvetica, sans-serif;">指向大端</span><span style="font-family: Arial, Helvetica, sans-serif;">；pSmall:指向小端；</span>
* 初始化：
* 1. 排序
* 2. pSmall指向小端，pBig指向大端
* 算法过程：
* 1. if *pBig + *pSmall == target, return <pSmall, pBig>;
* 2. if *pBig + *pSmal > target, pBig--;
* 3. if *pBig + *pSmal < target, pSmall++;
* 4. if pBig < end && pSmall >= begin; then jmp 1; else return null;
*/

int length = nums.size();
vector<int> rst;
vector<int> nums_sort = nums;

// 初始化
sort(nums_sort.begin(), nums_sort.end());    // 排序
int pSmall = 0;
int pBig = length - 1;
while(pBig > pSmall)
{
if(target == nums_sort.at(pSmall) + nums_sort.at(pBig))
{
break;
}
if(target < nums_sort.at(pSmall) + nums_sort.at(pBig))
{
pBig--;
continue;
}
if(target > nums_sort.at(pSmall) + nums_sort.at(pBig))
{
pSmall++;
continue;
}
}

int num_small = nums_sort.at(pSmall);
int num_big = nums_sort.at(pBig);
// 找到原来nums对应的idx
for(int i = 0; i < length; i++)
{
if(nums.at(i) == num_small || nums.at(i) == num_big)
{
rst.push_back(i + 1);
}
}
return rst;
}
};