代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution { public: vector<int> plusOne(vector<int> &digits) { int i,k=0; int a[100]; for(i=digits.size()-1;i>=0;i--) a[k++]=digits[i]; int c=1; for(i=0;i<k;i++) { int sum=a[i]+c; a[i]=sum%10; c=...
Plus One.go中,如何处理进位问题? Leetcode Golang 66. Plus One.go 的时间复杂度是多少? 思路 注意处理有进位的情况 code 代码语言:javascript 代码运行次数:0 运行 AI代码解释 func plusOne(digits []int) []int { l := len(digits) for i := l - 1; i >= 0; i-- { if digits[i] < 9...
Java实现 1classSolution {2publicListNode plusOne(ListNode head) {3ListNode dummy =newListNode(0);4dummy.next =head;5ListNode i =dummy;6ListNode j =dummy;7while(j.next !=null) {8j =j.next;9if(j.val != 9) {10i =j;11}12}13i.val++;14i =i.next;15while(i !=null) {16i.val...
代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution: def plusOne(self, digits): """ :type digits: List[int] :rtype: List[int] """ if digits[-1] != 9: # 如果各位不为9,则直接加1即可digits[-1] += 1 return digits else: i = len(digits) - 1 bit = 1 while i...
classSolution{publicint[] plusOne(int[] digits) {inti=digits.length -1; digits[i]++;intcarry=digits[i] /10; digits[i] = digits[i] %10;// 需要进位则一直向前while(carry !=0&& i >=1) { i--; digits[i] = digits[i] + carry; ...
One of them is marked in bold. Example 2: Input: N = 2, mines = [] Output: 1 Explanation: There is no plus sign of order 2, but there is of order 1. Example 3: Input: N = 1, mines = [[0, 0]] Output: 0 Explanation: There is no plus sign, so return 0. Note: ...
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classSolution {public: vector<int> plusOne(vector<int> &digits) {intend = digits.size() -1;for(inti = end;i>=0;--i){if(digits[i] ==9) digits[i] =0;else{ digits[i]+=1;returndigits; } } digits.insert(digits.begin(),1);returndigits; ...