Question

Source

• lintcode: (51) Previous Permuation

Problem

Given a list of integers, which denote a permutation.

Find the previous permutation in ascending order.

Example

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

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

Note

The list may contains duplicate integers.

题解

和前一题 Next Permutation 非常类似，这里找上一个排列，仍然使用字典序算法，大致步骤如下：

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] 小的数 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。

为何不从前往后呢？因为只有从后往前才能保证得到的是相邻的排列，可以举个实际例子自行分析。

Python

class Solution:
  # @param num :  a list of integer
  # @return : a list of integer
  def previousPermuation(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 previous permuation
   */
  vector<int> previousPermuation(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 minimum
        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: A list of integers
   * @return: A list of integers that's previous permuation
   */
  public ArrayList<Integer> previousPermuation(ArrayList<Integer> nums) {
    ArrayList<Integer> perm = new ArrayList<Integer>(nums);
    if (nums == null || nums.size() == 0) return perm;

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

    // step2: search the first perm[k] > perm[l] backward
    int l = perm.size() - 1;
    while (l > k && perm.get(l) >= perm.get(k)) {
      l--;
    }

    // step3: swap perm[k] with perm[l]
    Collections.swap(perm, k, l);

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

    return perm;
  }

  private void reverse(List<Integer> nums, int lb, int ub) {
    for (int i = lb, j = ub; i < j; i++, j--) {
      Collections.swap(nums, i, j);
    }
  }
}

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大的序列。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。

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)O(n), 使用了 temp 临时变量，空间复杂度可认为是 O(1)O(1)O(1).