twosum
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.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
給定兩個非空的鍊結串列，分別代表兩個非負的整數串列，串列內每個元素皆由反方向儲存(越前面的節點位數越低)，每一個節點代表一個位元。
將此兩個鍊結串列相加並回傳。
Example:
Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
參考答案：
int* twoSum(int* nums, int numsSize, int target){
int *ans = malloc(sizeof(int)*2);
for(int i=0;i<numsSize;i++){
for(int j=0;j<numsSize;j++){
if(target == (nums[i] + nums[j]) && i!=j ){
ans[0] = i,ans[1] = j;
return ans;
}
}
}
return ans;
}
學習目的：
學習兩個以上的回傳值型態宣告方法 malloc
c++ define:
void* malloc (size_t size);
//example
int *data = malloc(1000, sizeof(int)); //宣告
free(data); //釋放
測試程式:
//測試程式
int* twoSum(int* nums, int numsSize, int target){
int *ans = malloc(sizeof(int)*2);
for(int i=0;i<numsSize;i++){
for(int j=0;j<numsSize;j++){
if(target == (nums[i] + nums[j]) && i!=j ){
ans[0] = i,ans[1] = j;
return ans;
}
}
}
return ans;
}
int main(){
int nums[] = {2, 7, 11, 15};
int target = 9;
printf("sizeof(nums) = %1lu \n", sizeof(nums)/sizeof(int));
int *out = twoSum(nums, sizeof(nums)/sizeof(int), target);
printf("\n%d\n",out[0]);
printf("%d\n",out[1]);
return 0;
}
