leetcode问题:PlusOne 给定一个由整数组成的非空数组所表示的非负整数,在该数的基础上加一。 最高位数字存放在数组的首位, 数组中每个元素只存储一个数字。 你可以假设除了整数 0 之外,这个整数不会以零开头。 示例1: 输入:[1,2,3]输出:[1,2,4]解释:输入数组表示数字 123。 示例2: 输入:[4,3,2,1]...
1publicint[] plusOne(int[] digits) {2int[] result =newint[digits.length + 1];3for(inti = digits.length - 1; i >= 0; i--){4intdigit =digits[i];5if(digit < 9){6digits[i]++;7returndigits;8}else{9digits[i] = 0;10result[i+1] =digits[i];11}12}13result[0] = 1;14...
Can you solve this real interview question? Plus One - You are given a large integer represented as an integer array digits, where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-t
leetcode 66. Plus One Given anon-emptyarray of digits representing a non-negative integer, plus one to the integer. The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit. You may assume the integer do...
public class Solution { public int[] plusOne(int[] digits) { int carries = 1; for(int i = digits.length-1; i>=0 && carries > 0; i--){ // fast break when carries equals zero int sum = digits[i] + 1; digits[i] = sum % 10; ...
Plus One 【题目】Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. 【解答】在从低位到高位循环的时候,不需要单独设置一个布尔变量 carry 表示是否有进位,因为逢 9 就要...
C++ primer plus 过了一遍，今天做了 LeetCode 前几题，感觉有点吃力，得在草稿上写的明明白白才能理...
CompileC++files using command: g++ -std=c++11 -Wall src/bar.cpp -o bar OR You can build all the files usingmake(Use MinGW GCC and GNU Make on Windows). The☢means that you need to have a LeetCode Premium Subscription. ProblemSolution ...
369.Plus-One-Linked-List (M) 430.Flatten-a-Multilevel-Doubly-Linked-List (H-) 457.Circular-Array-Loop (H-) 708.Insert-into-a-Cyclic-Sorted-List (H-) 1474.Delete-N-Nodes-After-M-Nodes-of-a-Linked-List (M+) 1670.Design-Front-Middle-Back-Queue (M+) 1756.Design-Most-Recently-Used...
*/varplusOne=function(digits){for(leti=digits.length-1;i>=0;i--){if(digits[i]<9){digits[i]=digits[i]+1returndigits}else{digits[i]=0}}// 注意:这句unshift是为了测试用例[9]/[9,9,9]这种列出的情况digits.unshift(1)returndigits};...