31. Next Permutation

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.

从数组末尾开始遍历第i个元素。
如果后一项小于前一项,则排序关系正确。
反之则将i与遍历过的部分中比i大的第一个数字交换。
然后对已遍历的部分排序。

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class Solution {
public void nextPermutation(int[] nums) {
int flag = 0; //标记,如果没有下一个排列时,排序数组。
if (nums.length != 1){
int i = nums.length -2;
while (i >= 0){
if (nums[i + 1] <= nums[i]) { //从尾部开始,比较元素是否是大到小
i--;
continue;
}
else { //排序关系不正确时
for (int j = nums.length-1;j>i;j--){
if (nums[j] <= nums[i]){
continue;
}
int temp = nums[j]; //将i元素和遍历过的元素中第一个比nums[i]大的交换。
nums[j] = nums[i];
nums[i] = temp;
Arrays.sort(nums,i+1,nums.length); //排序i之后的数组。
flag = 1;
break;
}
break;
}
}
if (flag == 0 ){ //如果全部从大到小,则排序整个数组。
Arrays.sort(nums);
}
}
}
}

Author

Xander

Posted on

2022-04-04

Updated on

2022-04-20

Licensed under

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