[leetcode]1.Two Sum

Two Sum

描述：
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：进行两次for循环实现数组中任意两数相加求和，并与目标值进行判断，输出对应的元素位置。

问题：一开始，在本地编译器编译可以通过（用给出的个例时），但submit时报错提示“load of null pointer of type ‘const int’ ”，整数类型空指针的加载。之后修改了line12中n的值，一开始错想成有Cn(n-1)种结果，即给n赋了该值，但n是for循环的边界，应该是数组中的个数，修改后即可accepted。

代码：

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target) {
    
    int *p; 
    p=(int *)malloc(sizeof(int)*2);    
     if(p!=NULL)  
    {   
         int i;
    int j;  
    int n=numsSize;

    if(n<0 || nums==NULL)
        return NULL;
    for(int i=1;i