31-Next Premutation

【题目】

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

【analyze】

1.从后往前找（找到第一个元素的值比后序的元素值 小，索引记为pivot）

2.再从后往前找（找到第一个元素的值比pivot处的元素大的值（不要=），索引记为large）,交换pivot和large

3.将pivot+1-end之间的元素逆转

【算法】

public class Solution {
    public void nextPermutation(int[] nums) {
        int len=nums.length;
        int pivot=len-1;
        while(pivot>0) {
            if(nums[pivot]>nums[pivot-1])
                break;
            pivot--;
        }
        if(pivot>0) {
            pivot--;
            int large=len-1;
            while(nums[pivot]>=nums[large])   //找出比pivot元素值大的large
                large--;
            swap(nums,pivot,large);
            reverse(nums,pivot+1,len-1);
        }else {
            reverse(nums,0,len-1);
        }
    }
    //将数据区间的元素逆转
    public void reverse(int[] nums,int first,int end) {
        for(int i=0;i<(end-first+1)/2;i++) {
            swap(nums,first+i,end-i);
        }
    }
    //交换元素
    public void swap(int[] nums,int a,int b) {
        int temp=nums[a];
        nums[a]=nums[b];
        nums[b]=temp;
    }
}