【LeetCode】66. Plus One
Difficulty: Easy
More:【目录】LeetCode Java实现
Description
https://leetcode.com/problems/plus-one/
Given a non-empty array 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 does not contain any leading zero, except the number 0 itself.
Example 1:
Input: [1,2,3] Output: [1,2,4] Explanation: The array represents the integer 123.
Example 2:
Input: [4,3,2,1] Output: [4,3,2,2] Explanation: The array represents the integer 4321.
Intuition
读完题,首先要考虑到的是:1. 负数情况;2. 溢出情况;3.数字的顺序 123是[1,2,3]存放还是[3,2,1]存放。
本题比较简单,只要根据当前位的数字是否小于9来判断是否进位。详见代码。
Solution
public int[] plusOne(int[] digits) { if(digits==null) return digits; for(int i=digits.length-1; i>=0; i--){ if(digits[i] < 9){ digits[i]++; return digits; } digits[i]=0; } int[] newDigits=new int[digits.length+1]; newDigits[0]=1; //其余位的数都是0,如:999+1=1000 return newDigits; }
What I've learned
这道题自己一开始想的是使用递归函数,传入当前的index,后面参考别人的代码才发现,其实只要遍历数组就可以了。
如果最高位还进位的话,那么说明最高位是1,其余位都是0。其余位不用从旧的数组赋值到新的数组中了。
More:【目录】LeetCode Java实现