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feat: add solutions to lc problem: No.1053
No.1053.Previous Permutation With One Swap
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‎solution/1000-1099/1053.Previous Permutation With One Swap/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心**
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我们先从右到左遍历数组，找到第一个满足 `arr[i - 1] > arr[i]` 的下标 `i`，此时 `arr[i - 1]` 就是我们要交换的数字，我们再从右到左遍历数组，找到第一个满足 `arr[j] < arr[i - 1]` 且 `arr[j] != arr[j - 1]` 的下标 `j`，此时我们交换 `arr[i - 1]` 和 `arr[j]` 后返回即可。
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如果遍历完数组都没有找到满足条件的下标 `i`，说明数组已经是最小排列，直接返回原数组即可。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def prevPermOpt1(self, arr: List[int]) -> List[int]:
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n = len(arr)
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for i in range(n - 1, 0, -1):
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if arr[i - 1] > arr[i]:
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for j in range(n - 1, i - 1, -1):
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if arr[j] < arr[i - 1] and arr[j] != arr[j - 1]:
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arr[i - 1], arr[j] = arr[j], arr[i - 1]
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return arr
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return arr
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] prevPermOpt1(int[] arr) {
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int n = arr.length;
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for (int i = n - 1; i > 0; --i) {
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if (arr[i - 1] > arr[i]) {
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for (int j = n - 1; j > i - 1; --j) {
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if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
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int t = arr[i - 1];
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arr[i - 1] = arr[j];
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arr[j] = t;
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return arr;
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}
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}
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}
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}
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return arr;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> prevPermOpt1(vector<int>& arr) {
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int n = arr.size();
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for (int i = n - 1; i > 0; --i) {
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if (arr[i - 1] > arr[i]) {
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for (int j = n - 1; j > i - 1; --j) {
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if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
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swap(arr[i - 1], arr[j]);
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return arr;
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}
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}
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}
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}
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return arr;
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}
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};
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```
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### **Go**
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```go
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func prevPermOpt1(arr []int) []int {
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n := len(arr)
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for i := n - 1; i > 0; i-- {
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if arr[i-1] > arr[i] {
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for j := n - 1; j > i-1; j-- {
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if arr[j] < arr[i-1] && arr[j] != arr[j-1] {
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arr[i-1], arr[j] = arr[j], arr[i-1]
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return arr
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}
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}
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}
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}
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return arr
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}
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```
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### **...**

‎solution/1000-1099/1053.Previous Permutation With One Swap/README_EN.md

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### **Python3**
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```python
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class Solution:
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def prevPermOpt1(self, arr: List[int]) -> List[int]:
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n = len(arr)
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for i in range(n - 1, 0, -1):
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if arr[i - 1] > arr[i]:
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for j in range(n - 1, i - 1, -1):
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if arr[j] < arr[i - 1] and arr[j] != arr[j - 1]:
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arr[i - 1], arr[j] = arr[j], arr[i - 1]
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return arr
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return arr
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```
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### **Java**
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```java
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class Solution {
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public int[] prevPermOpt1(int[] arr) {
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int n = arr.length;
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for (int i = n - 1; i > 0; --i) {
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if (arr[i - 1] > arr[i]) {
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for (int j = n - 1; j > i - 1; --j) {
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if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
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int t = arr[i - 1];
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arr[i - 1] = arr[j];
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arr[j] = t;
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return arr;
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}
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}
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}
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}
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return arr;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> prevPermOpt1(vector<int>& arr) {
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int n = arr.size();
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for (int i = n - 1; i > 0; --i) {
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if (arr[i - 1] > arr[i]) {
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for (int j = n - 1; j > i - 1; --j) {
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if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
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swap(arr[i - 1], arr[j]);
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return arr;
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}
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}
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}
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}
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return arr;
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}
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};
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```
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### **Go**
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```go
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func prevPermOpt1(arr []int) []int {
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n := len(arr)
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for i := n - 1; i > 0; i-- {
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if arr[i-1] > arr[i] {
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for j := n - 1; j > i-1; j-- {
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if arr[j] < arr[i-1] && arr[j] != arr[j-1] {
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arr[i-1], arr[j] = arr[j], arr[i-1]
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return arr
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}
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}
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}
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}
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return arr
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}
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```
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### **...**

‎solution/1000-1099/1053.Previous Permutation With One Swap/Solution.cpp

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class Solution {
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public:
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vector<int> prevPermOpt1(vector<int>& arr) {
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int n = arr.size();
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for (int i = n - 1; i > 0; --i) {
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if (arr[i - 1] > arr[i]) {
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for (int j = n - 1; j > i - 1; --j) {
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if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
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swap(arr[i - 1], arr[j]);
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return arr;
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}
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}
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}
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}
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return arr;
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}
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};

‎solution/1000-1099/1053.Previous Permutation With One Swap/Solution.go

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func prevPermOpt1(arr []int) []int {
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n := len(arr)
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for i := n - 1; i > 0; i-- {
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if arr[i-1] > arr[i] {
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for j := n - 1; j > i-1; j-- {
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if arr[j] < arr[i-1] && arr[j] != arr[j-1] {
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arr[i-1], arr[j] = arr[j], arr[i-1]
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return arr
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}
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}
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}
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}
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return arr
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}

‎solution/1000-1099/1053.Previous Permutation With One Swap/Solution.java

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class Solution {
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public int[] prevPermOpt1(int[] A) {
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for (int i = A.length - 2; i >= 0; --i) {
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if (A[i] > A[i + 1]) {
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int k = i + 1;
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for (int j = k + 1; j < A.length; ++j) {
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if (A[j] < A[i] && A[j] > A[k]) {
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k = j;
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public int[] prevPermOpt1(int[] arr) {
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int n = arr.length;
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for (int i = n - 1; i > 0; --i) {
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if (arr[i - 1] > arr[i]) {
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for (int j = n - 1; j > i - 1; --j) {
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if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
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int t = arr[i - 1];
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arr[i - 1] = arr[j];
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arr[j] = t;
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return arr;
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}
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}
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int t = A[i];
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A[i] = A[k];
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A[k] = t;
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return A;
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}
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}
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return A;
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return arr;
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}
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}
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}

‎solution/1000-1099/1053.Previous Permutation With One Swap/Solution.py

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class Solution:
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def prevPermOpt1(self, arr: List[int]) -> List[int]:
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n = len(arr)
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for i in range(n - 1, 0, -1):
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if arr[i - 1] > arr[i]:
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for j in range(n - 1, i - 1, -1):
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if arr[j] < arr[i - 1] and arr[j] != arr[j - 1]:
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arr[i - 1], arr[j] = arr[j], arr[i - 1]
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return arr
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return arr

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