2021-07-23 15:44:19 +00:00
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> 如果阅读时,发现错误,或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ,备注 github + 题目 + 问题 向我反馈
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2021-03-20 08:30:29 +00:00
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>
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> 感谢支持,该仓库会一直维护,希望对各位有一丢丢帮助。
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>
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2021-07-27 18:26:32 +00:00
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号:袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步,想要和题友一起刷题,互相监督的同学,可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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2021-03-20 08:30:29 +00:00
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2021-03-20 07:48:03 +00:00
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#### [66. 加一](https://leetcode-cn.com/problems/plus-one/)
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2021-03-17 11:49:19 +00:00
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**题目描述**
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> 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。
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>
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> 最高位数字存放在数组的首位, 数组中每个元素只存储单个数字。
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>
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> 你可以假设除了整数 0 之外,这个整数不会以零开头。
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**示例 1:**
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> 输入:digits = [1,2,3]
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> 输出:[1,2,4]
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> 解释:输入数组表示数字 123。
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**示例 2:**
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> 输入:digits = [4,3,2,1]
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> 输出:[4,3,2,2]
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> 解释:输入数组表示数字 4321。
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**示例 3:**
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输入:digits = [0]
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输出:[1]
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**数组遍历**
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**题目解析**
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我们思考一下,加一的情况一共有几种情况呢?是不是有以下三种情况
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![加一](https://cdn.jsdelivr.net/gh/tan45du/github.io.phonto2@master/myphoto/加一.3lp9zidw61s0.png)
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则我们根据什么来判断属于第几种情况呢?
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2021-07-23 15:44:19 +00:00
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我们可以根据当前位 余 10 来判断,这样我们就可以区分属于第几种情况了,大家直接看代码吧,很容易理解的。
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2021-03-17 11:49:19 +00:00
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2021-07-10 04:20:02 +00:00
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Java Code:
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2021-03-17 11:49:19 +00:00
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```java
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class Solution {
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public int[] plusOne(int[] digits) {
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//获取长度
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int len = digits.length;
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for (int i = len-1; i >= 0; i--) {
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digits[i] = (digits[i] + 1) % 10;
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//第一种和第二种情况,如果此时某一位不为 0 ,则直接返回即可。
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if (digits[i] != 0) {
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return digits;
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}
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2021-07-27 18:26:32 +00:00
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2021-03-17 11:49:19 +00:00
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}
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//第三种情况,因为数组初始化每一位都为0,我们只需将首位设为1即可
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2021-07-27 18:26:32 +00:00
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int[] arr = new int[len+1];
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2021-03-17 11:49:19 +00:00
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arr[0] = 1;
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return arr;
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}
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}
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```
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2021-07-10 04:20:02 +00:00
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Python Code:
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```python
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from typing import List
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class Solution:
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def plusOne(self, digits: List[int])->List[int]:
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# 获取长度
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leng = len(digits)
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for i in range(leng - 1, -1, -1):
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digits[i] = (digits[i] + 1) % 10
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# 第一种和第二种情况,如果此时某一位不为 0 ,则直接返回即可。
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if digits[i] != 0:
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return digits
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# 第三种情况,因为数组初始化每一位都为0,我们只需将首位设为1即可
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arr = [0] * (leng + 1)
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arr[0] = 1
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return arr
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```
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2021-07-17 04:13:15 +00:00
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2021-07-15 15:14:17 +00:00
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C++ Code:
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```cpp
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class Solution {
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public:
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vector<int> plusOne(vector<int>& digits) {
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for(int i = digits.size() - 1; i >= 0; --i){
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digits[i] = (digits[i] + 1)%10;
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if(digits[i]) return digits;
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}
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for(int & x: digits) x = 0;
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digits.emplace_back(1);
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reverse(digits.begin(), digits.end());
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return digits;
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}
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};
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```
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2021-07-17 04:13:15 +00:00
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Swift Code:
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```swift
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class Solution {
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func plusOne(_ digits: [Int]) -> [Int] {
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let count = digits.count
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var digits = digits
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for i in stride(from: count - 1, through: 0, by: -1) {
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digits[i] = (digits[i] + 1) % 10
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if digits[i] != 0 {
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return digits
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}
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}
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var arr: [Int] = Array.init(repeating: 0, count: count + 1)
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arr[0] = 1
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return arr
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}
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}
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```
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2021-07-27 18:26:32 +00:00
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Go Code:
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```go
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func plusOne(digits []int) []int {
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l := len(digits)
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for i := l - 1; i >= 0; i-- {
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digits[i] = (digits[i] + 1) % 10
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if digits[i] != 0 {
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return digits
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}
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}
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digits = append([]int{1}, digits...)
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return digits
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}
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```
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