My LeetCode first day

題目

我的思路

看到這題首先想到的就是直接遍歷所有元素,簡單暴力且不會出錯

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target, int* returnSize){
    int* rec=(int*)malloc(sizeof(int)*2);
    int i,j;
for(i=0;i<numsSize-1;i++)
{
    for(j=i+1;j<numsSize;j++)
    {
if(nums[i]+nums[j]==target)
{
rec[0]=i;
rec[1]=j;
break;
}
    }
}
* returnSize=2;
return rec;
}

結果

優化解法二分內二分

int bisect_right(int *nums, int *index, int left, int right, int target, bool isright) {
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[index[mid]] > target) {
            right = mid;
        } else if (nums[index[mid]] == target) {
            return mid;
        } else left = mid + 1;
    }
    if (isright) return left - 1;  // 沒有退出就是左邊小一，否則就是left情況
    return left;
}

int* twoSum(int* nums, int numsSize, int target, int* returnSize){
    int *index = (int *)malloc(sizeof(int) * numsSize);
    for (int i = 0; i < numsSize; i++) index[i] = i;

    int cmp(int *a, int *b) {
        return nums[*a] - nums[*b];
    }
    qsort(index, numsSize, sizeof(index[0]), cmp);


    int left = 0, right = numsSize - 1;
    while (left < right) {  // 閉區間，因爲存在[2,3,3],target=6
        if (nums[index[left]] + nums[index[right]] < target) {
            // 比target小，因爲index升序，因此一定是左邊小！
            left = bisect_right(nums, index, left, right, target - nums[index[right]], false);
        } else if (nums[index[left]] + nums[index[right]] > target) {
            // 這時有可能死循環，[1,2,5,7,9],target=9,上界始終是9
            right = bisect_right(nums, index, left, right, target - nums[index[left]], true);
        } else {
            int *res = (int *)malloc(sizeof(int) * 2);
            res[0] = index[left];
            res[1] = index[right];
            *returnSize = 2;
            return res;
        }
    }
    return NULL;
}

每天進步一點點,每天收穫一點點.