代碼如下：Return an array of size *returnSize.Note: The returned array must be malloced, assume caller calls free()./int twoSum(int* numbers, int numbersSize, int target, int* returnSize) {int i=0;int j=numbersSize-1;while(numbers[i]+numbers[j]!=target){if(numbers[i]+numbers[j]>target){j=j-1;}else{i=i+1;}}returnSize[0]=i+1;returnSize[1]=j+1;return returnSize;}運(yùn)行的時(shí)候出不來結(jié)果啊，請(qǐng)問這是為什么，我該怎么改呢？
1 回答
波斯汪
TA貢獻(xiàn)1811條經(jīng)驗(yàn) 獲得超4個(gè)贊
不考慮你程序本身的對(duì)錯(cuò)，僅僅從輸入輸出的角度來說，你返回的是結(jié)果數(shù)組的指針，而returnsize的作用是返回，告知調(diào)用者結(jié)果數(shù)組的大小
/* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* numbers, int numbersSize, int target, int * returnSize) {
int i=0;
int j=numbersSize-1;
while(numbers[i]+numbers[j]!=target)
{
if(numbers[i]+numbers[j]>target)
{
j=j-1;
}
else
{
i=i+1;
}
}
*returnSize = 2;
int * r = new int[2];
r[0]=i+1;
r[1]=j+1;
return r;
}
int _tmain(int argc, _TCHAR* argv[])
{
int a[] = {2,7,15,9};
int t = 9;
int rs;
int * r = twoSum(a, 4, t, &rs);
for (int i = 0; i < rs; i++)
printf("%d ", r[i]);
return 0;
}應(yīng)該是這么一個(gè)邏輯
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