Add solution #31

Author: JoshCrozier

README.md

@@ -7,6 +7,7 @@
 |#|Title|Difficulty|
 |:---|:---|:---|
 4|[Median of Two Sorted Arrays](./0004-median-of-two-sorted-arrays.js)|Hard|
+31|[Next Permutation](./0031-next-permutation.js)|Medium|
 36|[Valid Sudoku](./0036-valid-sudoku.js)|Medium|
 43|[Multiply Strings](./0043-multiply-strings.js)|Medium|
 58|[Length of Last Word](./0058-length-of-last-word.js)|Easy|
@@ -76,4 +77,4 @@
 
 [MIT License](https://opensource.org/licenses/MIT)
 
-Copyright (c) 2019-2020 [Josh Crozier](https://joshcrozier.com)
+Copyright (c) 2019-2021 [Josh Crozier](https://joshcrozier.com)

solutions/0031-next-permutation.js

@@ -0,0 +1,50 @@
+/**
+ * 31. Next Permutation
+ * https://leetcode.com/problems/next-permutation/
+ * Difficulty: Medium
+ *
+ * Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
+ *
+ * If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).
+ *
+ * The replacement must be in place and use only constant extra memory.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {void} Do not return anything, modify nums in-place instead.
+ */
+ var nextPermutation = function(nums) {
+  let i = nums.length - 2;
+
+  while (i >= 0 && nums[i + 1] <= nums[i]) {
+    i--;
+  }
+
+  if (i >= 0) {
+    let cursor = nums.length - 1;
+    while (nums[cursor] <= nums[i]) {
+      cursor--;
+    }
+    swap(nums, i, cursor);
+  }
+
+  reverse(nums, i + 1);
+};
+
+function reverse(nums, start) {
+  let i = start;
+  let j = nums.length - 1;
+
+  while (i < j) {
+    swap(nums, i, j);
+    i++;
+    j--;
+  }
+}
+
+function swap(nums, i, j) {
+  const temp = nums[i];
+  nums[i] = nums[j];
+  nums[j] = temp;
+}