66. Plus One 数组加1
[抄题]:
Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.
You may assume the integer do not contain any leading zero, except the number 0 itself.
The digits are stored such that the most significant digit is at the head of the list.
给定 [1,2,3]
表示 123, 返回 [1,2,4]
.
给定 [9,9,9]
表示 999, 返回 [1,0,0,0]
.
[暴力解法]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
不知道对进不进位如何分类。提前return就无忧了
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 不需要对answer数组的后面若干位赋值了,初始化时自动=0。感觉是针对此题特殊的
- 有角标循环的时候,提前备注:0- n-1,无论正序、倒序
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
备注0 to n -1
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
for (int i = n - 1; i >= 0; i--) { if (digits[i] < 9) { digits[i]++; return digits; }else { digits[i] = 0; } }
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
369. Plus One Linked List dummynode要想到,两根指针也行 我还是太天真
[代码风格] :
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
public class Solution { /** * @param digits: a number represented as an array of digits * @return: the result */ public int[] plusOne(int[] digits) { //not carry int n = digits.length; //0 to n-1 whenever for (int i = n - 1; i >= 0; i--) { if (digits[i] < 9) { digits[i]++; return digits; }else { digits[i] = 0; } } //carry, need new array int[] answer = new int[n + 1]; answer[0] = 1; return answer; } }