2023每日刷题（十一）

Leetcode—66.加一

实现代码1

/*** Note: The returned array must be malloced, assume caller calls free().*/
int* plusOne(int* digits, int digitsSize, int* returnSize){int num = 0;int i = 0;int arr[110] = {0};// 进位标识符int carryOver = 0;int cnt = digitsSize - 1;int tmp = digits[cnt] + 1;if(tmp == 10) {carryOver = 1;arr[cnt] = 0;} else {arr[cnt] = tmp;}int cur = cnt;while(cnt--) {if(carryOver == 0) {arr[--cur] = digits[cnt];} else {tmp = digits[cnt] + 1;if(tmp == 10) {carryOver = 1;arr[--cur] = 0;} else {carryOver = 0;arr[--cur] = tmp;}}}if(carryOver) {*returnSize = digitsSize + 1;} else {*returnSize = digitsSize;}cur = *returnSize;int *arrList = (int *)malloc((cur) * sizeof(int));while(digitsSize--) {arrList[--cur] = arr[digitsSize];}if(carryOver) {arrList[0] = 1;}return arrList;
}

运行结果

实现代码2

/*** Note: The returned array must be malloced, assume caller calls free().*/
int* plusOne(int* digits, int digitsSize, int* returnSize){int cnt = 0;int i;for(i = 0; i < digitsSize; i++) {if(digits[i] == 9) {cnt++;}}// 全部元素都为9if(cnt == digitsSize) {int *ans = (int *)malloc((digitsSize + 1) * sizeof(int));ans[0] = 1;int j;for(j = 1; j < digitsSize + 1; j++) {ans[j] = 0;}*returnSize = digitsSize + 1;return ans;} else {int *ans = (int *)malloc(digitsSize * sizeof(int));int c = 0;int i;for(i = digitsSize - 1; i >= 0; i--) {if(i == digitsSize - 1) {ans[i] = digits[i] + 1;c = ans[i] / 10;ans[i] = ans[i] % 10;} else {ans[i] = digits[i] + c;c = ans[i] / 10;ans[i] = ans[i] % 10;}}*returnSize = digitsSize;return ans;}    
}

运行结果

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