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Leetcode—66.加一【简单】

article 2024/11/17 13:41:43

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9
    if(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;
    }    
}