下一个排列

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int i = nums.size() - 2;
        while (i >= 0 && nums[i] >= nums[i + 1]) {
            i--;
        }
        if (i >= 0) {
            int j = nums.size() - 1;
            while (j >= 0 && nums[i] >= nums[j]) {
                j--;
            }
            swap(nums[i], nums[j]);
        }
        reverse(nums.begin() + i + 1, nums.end());
    }
};
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/next-permutation/solution/xia-yi-ge-pai-lie-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

class Solution {
    public void nextPermutation(int[] nums) {
        if (nums == null || nums.length == 0) return;
        int firstIndex = -1;
        for (int i = nums.length - 2; i >= 0; i--) {
            if (nums[i] < nums[i + 1]) {
                firstIndex = i;
                break;
            }
        }
        if (firstIndex == -1) {
            reverse(nums, 0, nums.length - 1);
            return;
        }
        int secondIndex = -1;
        for (int i = nums.length - 1; i >= 0; i--) {
            if (nums[i] > nums[firstIndex]) {
                secondIndex = i;
                break;
            }
        }
        swap(nums, firstIndex, secondIndex);
        reverse(nums, firstIndex + 1, nums.length - 1);
        return;
    }
    private void reverse(int[] nums, int i, int j) {
        while (i < j) {
            swap(nums, i++, j--);
        }
    }
    private void swap(int[] nums, int i, int i1) {
        int tmp = nums[i];
        nums[i] = nums[i1];
        nums[i1] = tmp;
    }
}
作者：powcai
链接：https://leetcode-cn.com/problems/next-permutation/solution/xia-yi-ge-pai-lie-by-powcai/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。