題目敘述
A permutation of an array of integers is an arrangement of its members into a sequence or linear order.
- For example, for
arr = [1,2,3], the following are considered permutations of arr: [1,2,3], [1,3,2], [3,1,2], [2,3,1].
The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).
- For example, the next permutation of
arr = [1,2,3] is [1,3,2].
- Similarly, the next permutation of
arr = [2,3,1] is [3,1,2].
- While the next permutation of
arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.
Given an array of integers nums, find the next permutation of nums.
The replacement must be in place and use only constant extra memory.
Example 1:
Input: nums = [1,2,3]
Output: [1,3,2]
Example 2:
Input: nums = [3,2,1]
Output: [1,2,3]
Example 3:
Input: nums = [1,1,5]
Output: [1,5,1]
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 100
題目翻譯
給一個整數陣列 nums，需要找出此排序之後的下一個大的排序。如果已經是最大的排序的話就升序輸出。
解法解析
- 從結尾往開頭尋找，找到第一個降冪的值
x
- 從第一步找到的值
x 的位置到結尾的這段資料中，找到大於 x 中的最小值 y (例如 x = 4, x-> n-1 中有 5, 6, 9，則選擇 5)
- 將
x 和 y 的位置互換
- 將
x-> n-1 的值做從小到大排序
解法範例
Go
func nextPermutation(nums []int) {
nlen := len(nums)
if nlen <= 1 {
return
}
for i := nlen - 1; i > 0; i-- {
if nums[i] <= nums[i-1] {
continue
}
j := i
for ; j < nlen-1; j++ {
if nums[j] > nums[i-1] && nums[j+1] <= nums[i-1] {
break
}
}
nums[i-1], nums[j] = nums[j], nums[i-1]
reverse(nums[i:])
return
}
reverse(nums)
}
func reverse(nums []int) {
for i, j := 0, len(nums)-1; i < j; i, j = i+1, j-1 {
nums[i], nums[j] = nums[j], nums[i]
}
}
JavaScript
var nextPermutation = function (nums) {
function swap(i, j) {
const temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
function reverse(start) {
let end = nums.length - 1;
while (start < end) {
swap(start, end);
start++;
end--;
}
}
let i = nums.length - 2;
while (i >= 0 && nums[i + 1] <= nums[i]) {
i--;
}
if (i >= 0) {
let j = nums.length - 1;
while (j >= 0 && nums[j] <= nums[i]) {
j--;
}
swap(i, j);
}
reverse(i + 1);
};
Kotlin
class Solution {
fun nextPermutation(nums: IntArray) {
var i = nums.size - 2
while (i >= 0 && nums[i + 1] <= nums[i]) {
i--
}
if (i >= 0) {
var j = nums.size - 1
while (j >= 0 && nums[j] <= nums[i]) {
j--
}
swap(nums, i, j)
}
reverse(nums, i + 1)
}
fun swap(nums: IntArray, i: Int, j: Int) {
val temp = nums[i]
nums[i] = nums[j]
nums[j] = temp
}
fun reverse(nums: IntArray, i: Int) {
var left = i
var right = nums.size - 1
while (left < right) {
swap(nums, left, right)
left++
right--
}
}
}
PHP
class Solution
{
function nextPermutation(&$nums)
{
$length = count($nums);
$i = $length - 2;
while ($i >= 0 && $nums[$i] >= $nums[$i + 1]) {
$i--;
}
if ($i >= 0) {
$j = $length - 1;
while ($nums[$i] >= $nums[$j]) {
$j--;
}
[$nums[$i], $nums[$j]] = [$nums[$j], $nums[$i]];
}
$revStart = $i + 1;
$revEnd = $length - 1;
while ($revStart < $revEnd) {
[$nums[$revStart], $nums[$revEnd]] = [$nums[$revEnd], $nums[$revStart]];
$revStart++;
$revEnd--;
}
}
}
Python
class Solution:
def nextPermutation(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
i = len(nums) - 2
while i >= 0 and nums[i] >= nums[i + 1]:
i -= 1
if i >= 0:
j = len(nums) - 1
while nums[i] >= nums[j]:
j -= 1
nums[i], nums[j] = nums[j], nums[i]
rev_start = i + 1
rev_end = len(nums) - 1
while rev_start < rev_end:
nums[rev_start], nums[rev_end] = nums[rev_end], nums[rev_start]
rev_start += 1
rev_end -= 1
Rust
Swift
class Solution {
func reverse(_ nums: inout [Int], _ left: Int, _ right: Int) {
var left = left
var right = right
while left < right {
nums.swapAt(left, right)
left += 1
right -= 1
}
}
func nextPermutation(_ nums: inout [Int]) {
guard nums.count > 1 else {
return
}
var idx1 = nums.count - 2
while idx1 >= 0 && nums[idx1] >= nums[idx1 + 1] {
idx1 -= 1
}
if idx1 >= 0 {
var idx2 = nums.count - 1
while idx2 >= 0 && nums[idx2] <= nums[idx1] {
idx2 -= 1
}
nums.swapAt(idx1, idx2)
}
reverse(&nums, idx1 + 1, nums.count - 1)
}
}