two pointer

31. Next Permutation Medium

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 <= 100
• 0 <= nums[i] <= 100

分析：

case1:
1,3,2,5,4

从右向左找到第一个顺序对，i点
1,3,2,5,4
    - -
    i
    
从i+1点向右找到大于nums[i]的最小值的位置
1,3,2,5,4
    -   -
    i   j

交换i，j的位置
1,3,4,5,2
    -   -
    i   j

对i之后的元素进行排序
1,3,4,2,5

case2:
4,3,2,1
没有找到逆序对，说明是permutation的最后一个，那么下一个就是permutation第一个

class Solution {
    public void nextPermutation(int[] nums) {
        if(nums.length==1) return ;
        //从右向左找第一个顺序对
        int i = nums.length-2;
        while(i>=0){
            if(nums[i+1]>nums[i]) break;
            i--;
        }
        //从当前i向右找到大于它的最小元素
        int j=i+1;
        if(i>=0){
            for(int k = i+1;k<nums.length;k++){
                if(nums[i]<nums[k] && nums[k]<nums[j]) j=k;
            }
        }
        //交换位置
        if(i>=0) swap(nums,i,j);
        //对于后续进行排序
        Arrays.sort(nums,i+1,nums.length);
    }
    public void swap(int[] nums, int a, int b){
        int temp = nums[a];
        nums[a] = nums[b];
        nums[b] = temp;
    }
}

 

标签：arr,nums,int,two,next,permutation,array,pointer
来源： https://www.cnblogs.com/cynrjy/p/16389815.html