力扣-31-下一个排列

链接

class Solution31 {
    public void nextPermutation(int[] nums) {
        if (nums == null || nums.length == 0) return;
        int len = nums.length;

        // 从后往前找到第一个递增对
        int sx = len - 1;
        for (int i = len - 2; i >= 0; i--) {
            if (nums[i] < nums[i+1]) {
                sx = i;
                break;
            }
        }
        if (sx == len-1) {    // 没有找到递增对，说明整个子串是递减的
            reverse(nums, 0);
        }
        // 在[sx+1, len)区间从后往前找第一个大于nums[sx]的位置
        int end = len - 1;
        for (int i = len - 1; i >= sx+1; i--) {
            if (nums[i] > nums[sx]) {
                end = i;
                break;
            }
        }

        // 交换nums[sx]和nums[end]
        swap(nums, sx, end);
        reverse(nums, sx+1);
    }

    public void swap(int[] nums, int x, int y) {
        int temp = nums[x];
        nums[x] = nums[y];
        nums[y] = temp;
    }

    public void reverse(int[] nums, int start) {
        int l = start;
        int r = nums.length - 1;
        while (l < r) {
            swap(nums, l, r);
            l++;
            r--;
        }
    }
}

 

简化一下：

class Solution31 {
    public void nextPermutation(int[] nums) {
        if (nums == null || nums.length == 0) return;
        int len = nums.length;

        int sx = len - 2;
        while (sx >= 0 && nums[sx] >= nums[sx+1]) sx--;
        if (sx >= 0) {
            int end = len-1;
            while (end >= sx+1 && nums[sx] >= nums[end]) end--;
            swap(nums, sx, end);
        }
        reverse(nums, sx+1);
    }

    public void swap(int[] nums, int x, int y) {
        int temp = nums[x];
        nums[x] = nums[y];
        nums[y] = temp;
    }

    public void reverse(int[] nums, int start) {
        int l = start;
        int r = nums.length - 1;
        while (l < r) {
            swap(nums, l, r);
            l++;
            r--;
        }
    }
}