leetcode第1题 ，Two Sum

题目：Two Sum
LeetCode:https://leetcode.com/problems/two-sum/description/

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.

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

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

思路：本题是给定一个未排序数组，和一个目标值，求解数组中两个值相加为该目标值的两个数据得到索引。
由于未进行排序，如果采用循环遍历，那么时间复杂度将会达到O（n×n).

int* twoSum1(int* nums,int numsSize,int target){    int i,j;    int* retArr = (int*)malloc(2 * sizeof(int));    if(nums == NULL)    {        return NULL;    }    if(numsSize < 2 )    {        return NULL;    }    int diff;    for(i = 0 ; i < numsSize - 1 ;i++)    {        diff = target - *(nums+i);        for(j = i+1; j < numsSize; j++ )        {            if(*(nums + j)  == diff )            {                *retArr = i;                *(retArr + 1) = j;                return retArr;            }        }    }}

如果给定的数组为排序递增数组，则使用以下方法。
求解的思路是：
比如数组为 [1,3,6,9,12,17];目标值为15。
设定两个索引：
int lowIndex,从;
lowIndex = 0;
hightIndex = numsSize - 1;

求解 arr[lowIndex] + arr[hightIndex]的和，与目标值进行比较。
比如：1+17 = 18 > 15;
则hightIndex–，1+12 = 13 < 15
则lowIndex++，3+12 = 15 == 15 ,因此，所求的数值极为3和12。

#include<stdio.h>#include<stdlib.h>int* twoSum(int* nums,int numsSize,int target);int* testArrNull;int  testArr[10] = {12,54,87,98,124,149,257,457,789,1234};int* targetArr;int main(int argc ,char *argv[]){    printf("\r\n********systerm start **************\r\n");    //test1 :  null nums    targetArr =  twoSum(testArrNull,10,36);    if(targetArr == NULL)    {        printf("test 1:error test\r\n");    }    else     {        printf("test 1 :value1 = %d, value2 = %d\r\n",*targetArr,*(targetArr + 1));    }    //test2 :  length < 2    targetArr =  twoSum(testArr,1,36);    if(targetArr == NULL)    {        printf("test 2: error test\r\n ");    }    else     {        printf("test 2: value1 = %d, value2 = %d\r\n",*targetArr,*(targetArr + 1));    }    //test3 : nums    targetArr =  twoSum(testArr,10,406);    if(targetArr == NULL)    {        printf("test 3:error test\r\n ");    }    else     {        printf("test 3 :value1 = %d, value2 = %d\r\n",*targetArr,*(targetArr + 1));    }    //test4 :  null nums    targetArr =  twoSum(testArr,10,544);    if(targetArr == NULL)    {        printf("test 4:error test\r\n ");    }    else     {        printf("test 4 :value1 = %d, value2 = %d\r\n",*targetArr,*(targetArr + 1));    }    //test5 :  null nums    targetArr =  twoSum(testArr,10,801);    if(targetArr == NULL)    {        printf("test 5:error test\r\n ");    }    else     {        printf("test 5 :value1 = %d, value2 = %d\r\n",*targetArr,*(targetArr + 1));    }    printf("\r\n********systerm finish **************\r\n");    return 1;}int* twoSum(int* nums,int numsSize,int target){    int lowIndex,hightIndex;    int index;    int targetTemp = 0;    int* retArr = (int*)malloc(2 * sizeof(int));    lowIndex = 0;     hightIndex = numsSize - 1;    if(nums == NULL)    {        return NULL;    }    if(numsSize < 2 )    {        return NULL;    }    for(index = 0; index < numsSize ; index ++)    {        targetTemp = nums[lowIndex] + nums[hightIndex];        if(targetTemp > target)        {            hightIndex--;        }        else if(targetTemp < target)        {            lowIndex++;        }        else         {            break;        }        if(hightIndex == lowIndex)        {            return NULL;        }        printf("lowIndex:%d, hightIndex:%d,targetTemp = %d,target = %d \r\n",lowIndex,hightIndex,targetTemp,target);    }    *retArr = lowIndex;    *(retArr + 1) = hightIndex;    return retArr;}