leetcode - 31.Next Permutation

最新推荐文章于 2020-12-07 13:42:30 发布

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最新推荐文章于 2020-12-07 13:42:30 发布

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分类专栏： leetcode题集文章标签： leetcode

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本文介绍了一种在原地将数组重新排列为字典序中下一个更大排列的算法实现。如果无法实现这样的排列，则将其重新排列为最小可能的顺序（即升序）。文章通过示例展示了如何使用该算法，并提供了完整的Java代码实现。

Next Permutation

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, do not allocate 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

Solution:

  public void nextPermutation(int[] num) {
    if (num == null || num.length < 2) {
      return;
    }
    int length = num.length;
    int index = length - 1;
    while (index > 0) {
      if (num[index - 1] < num[index])
        break;
      index--;
    }
    if (index == 0) {
      reverseSort(num, 0, length - 1);
    } else {
      int val = num[index - 1];
      int j = length - 1;
      while (j >= index) {
        if (num[j] > val)
          break;
        j--;
      }
      swap(num, j, index - 1);
      reverseSort(num, index, length - 1);
    }
  }

  public void swap(int[] num, int i, int j) {
    int temp = num[i];
    num[i] = num[j];
    num[j] = temp;
  }

  public void reverseSort(int[] num, int start, int end) {
    if (start > end)
      return;
    for (int i = start; i <= (end + start) / 2; i++)
      swap(num, i, start + end - i);
  }