LeeCode : 31. Next Permutation 下一全排列字典序排列

最新推荐文章于 2019-09-10 22:48:04 发布

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最新推荐文章于 2019-09-10 22:48:04 发布

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博客围绕实现下一个排列展开，要求将数字重新排列成字典序中下一个更大的排列，若无法实现则排成最小顺序，且需原地操作、仅用常量额外内存。给出了示例，还提及代码实现主要有找到小于、找到大于、尾部翻转三步。

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试题
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 and use only constant 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

代码
主要有三步：找到小于，找到大于，尾部翻转

class Solution {
    public void nextPermutation(int[] nums) {
        if(nums==null || nums.length<=1) return;
        int len = nums.length;
        
        int cur = len-1;
        while(cur-1>=0 && nums[cur-1]>=nums[cur]) cur--;
        if(cur==0){
            reverse(nums, 0, len-1);
            return;
        }
        
        cur--;
        int lo = cur;
        while(cur+1<len && nums[cur+1]>nums[lo]) cur++;
        swap(nums, lo, cur);
        reverse(nums, lo+1, len-1);
        return;
    }
    
    public void reverse(int[] nums, int left, int right){
        while(left<right){
            swap(nums, left, right);
            left++;
            right--;
        }
    }
    
    public void swap(int[] nums, int i, int j){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}