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Next Permutation

Source

Problem

Given a list of integers, which denote a permutation.

Find the next permutation in ascending order.

Example

For [1,3,2,3], the next permutation is [1,3,3,2]

For [4,3,2,1], the next permutation is [1,2,3,4]

Note

The list may contains duplicate integers.

题解

找下一个升序排列，C++ STL 源码剖析一书中有提及，Permutations 一小节中也有详细介绍，下面简要介绍一下字典序算法：

  1. 从后往前寻找索引满足 a[k] < a[k + 1], 如果此条件不满足，则说明已遍历到最后一个。
  2. 从后往前遍历，找到第一个比a[k]大的数a[l], 即a[k] < a[l].
  3. 交换a[k]与a[l].
  4. 反转k + 1 ~ n之间的元素。

由于这道题中规定对于[4,3,2,1], 输出为[1,2,3,4], 故在第一步稍加处理即可。

Python

class Solution:
    # @param num :  a list of integer
    # @return : a list of integer
    def nextPermutation(self, num):
        if num is None or len(num) <= 1:
            return num
        # step1: find nums[i] < nums[i + 1], Loop backwards
        i = 0
        for i in xrange(len(num) - 2, -1, -1):
            if num[i] < num[i + 1]:
                break
            elif i == 0:
                # reverse nums if reach maximum
                num = num[::-1]
                return num
        # step2: find nums[i] < nums[j], Loop backwards
        j = 0
        for j in xrange(len(num) - 1, i, -1):
            if num[i] < num[j]:
                break
        # step3: swap betwenn nums[i] and nums[j]
        num[i], num[j] = num[j], num[i]
        # step4: reverse between [i + 1, n - 1]
        num[i + 1:len(num)] = num[len(num) - 1:i:-1]

        return num

C++

class Solution {
public:
    /**
     * @param nums: An array of integers
     * @return: An array of integers that's next permuation
     */
    vector<int> nextPermutation(vector<int> &nums) {
        if (nums.empty() || nums.size() <= 1) {
            return nums;
        }
        // step1: find nums[i] < nums[i + 1]
        int i = 0;
        for (i = nums.size() - 2; i >= 0; --i) {
            if (nums[i] < nums[i + 1]) {
                break;
            } else if (0 == i) {
                // reverse nums if reach maximum
                reverse(nums, 0, nums.size() - 1);
                return nums;
            }
        }
        // step2: find nums[i] < nums[j]
        int j = 0;
        for (j = nums.size() - 1; j > i; --j) {
            if (nums[i] < nums[j]) break;
        }
        // step3: swap betwenn nums[i] and nums[j]
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
        // step4: reverse between [i + 1, n - 1]
        reverse(nums, i + 1, nums.size() - 1);

        return nums;

    }

private:
    void reverse(vector<int>& nums, int start, int end) {
        for (int i = start, j = end; i < j; ++i, --j) {
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
        }
    }
};

Java

public class Solution {
    /**
     * @param nums: an array of integers
     * @return: return nothing (void), do not return anything, modify nums in-place instead
     */
    public void nextPermutation(int[] nums) {
        if (nums == null || nums.length == 0) return;

        // step1: search the first nums[k] < nums[k+1] backward
        int k = -1;
        for (int i = nums.length - 2; i >= 0; i--) {
            if (nums[i] < nums[i + 1]) {
                k = i;
                break;
            }
        }
        // if current rank is the largest, reverse it to smallest, return
        if (k == -1) {
            reverse(nums, 0, nums.length - 1);
            return;
        }

        // step2: search the first nums[k] < nums[l] backward
        int l = nums.length - 1;
        while (l > k && nums[l] <= nums[k]) l--;

        // step3: swap nums[k] with nums[l]
        int temp = nums[k];
        nums[k] = nums[l];
        nums[l] = temp;

        // step4: reverse between k+1 and nums.length-1;
        reverse(nums, k + 1, nums.length - 1);
    }

    private void reverse(int[] nums, int lb, int ub) {
        for (int i = lb, j = ub; i < j; i++, j--) {
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
        }
    }
}

源码分析

和 Permutation 一小节类似，这里只需要注意在step 1中i == -1时需要反转之以获得最小的序列。对于有重复元素，只要在 step1和 step2中判断元素大小时不取等号即可。Lintcode 上给的注释要求（其实是 Leetcode 上的要求）和实际给出的输出不一样。

复杂度分析

最坏情况下，遍历两次原数组，反转一次数组，时间复杂度为 O(n)O(n), 使用了 temp 临时变量，空间复杂度可认为是 O(1)O(1).