很久没有做算法题,本来一道简单的题目,还是耗费了一些时间在debug上。
题目描述:
1. Two Sum
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].
我使用的是c语言编程,一开始使用了直接申请一个数组作为返回元素,结果出现了报错。
load of null pointer of type 'const int'
依据提示,使用malloc动态分配的方式,解决了报错。
我的代码如下,写的比较蠢,日后思考一些高级一点的
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target) {
int flag=0,i,j;
int *ans=(int*)malloc(2*sizeof(int));
for(i=0;i<=numsSize-2;i++){
for(j = i+1;j<=numsSize-1;j++)
{
if((nums[i]+nums[j])==target) {
flag=1;
break;
}
}
if(flag==1) break;
}
ans[0]=i;
ans[1]=j;
return ans;
}