31 Next Permutations【M】【19】

最新推荐文章于 2024-09-07 04:25:13 发布

原创最新推荐文章于 2024-09-07 04:25:13 发布 · 315 阅读

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本文介绍了一个算法问题：如何在原地将一组数字重新排列为字典序中下一个更大的排列。如果无法形成更大的排列，则将其变为最小的排列（升序）。文章通过示例说明了该问题，并提供了一种解决方案。

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

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算的还是挺麻烦的，用最naive的方法做的。。。理论上应该还有更棒的。。。。。。

class Solution(object):
    def nextPermutation(self, nums):

        t = 0

        a = nums[:]
        a.sort(reverse = True)
        if a[:] == nums[:]:
            nums.reverse()
            return

        l = len(nums)

        if l == 1:# or nums == nums[::-1]:
            return

        t = l - 1
        while t >= 0 and nums[t] > nums[t-1]:
            t -= 1

        if l - t > 1:
            nums[l-1],nums[l-2] = nums[l-2],nums[l-1]
            return

        if t == l - 1:
            while t > 0 and nums[t] <= nums[t-1]:
                t -= 1
            a = nums[t-1:l][::-1]
            a.sort()
            pos = a.index(nums[t-1],0,len(a)) + 1
            while a[pos] == a[pos - 1]:
                pos += 1
            nums[t-1] = a[pos]
            del a[pos]
            nums[t:l] = a[:]
            return