767. Reorganize String

最新推荐文章于 2025-02-03 15:37:10 发布

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最新推荐文章于 2025-02-03 15:37:10 发布

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分类专栏： leetcode 优先队列

本文链接：https://blog.youkuaiyun.com/zjucor/article/details/79119502

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本文介绍了一种解决特定字符串重组问题的算法。该问题要求输入一个字符串S，并检查是否可以通过重新排列字符来避免相邻字符相同。文章提供了两种解决方案，一种尝试通过贪心算法实现，但存在局限性；另一种使用优先队列实现，更为有效。

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Given a string S, check if the letters can be rearranged so that two characters that are adjacent to each other are not the same.

If possible, output any possible result. If not possible, return the empty string.

Example 1:

Input: S = "aab"
Output: "aba"

Example 2:

Input: S = "aaab"
Output: ""

Note:

S will consist of lowercase letters and have length in range [1, 500].

思路：

最开始想分奇数位，偶数位填充，按照出现次数从高到低填充，优先选用剩余空间的填，有点贪心的味道

from collections import Counter
class Solution:
    def reorganizeString(self, S):
        """
        :type S: str
        :rtype: str
        """
        if len(S) == 1: return S
        p = ['']*len(S)
        d = Counter(S)
        i, j = 0, 1
        most = d.most_common(1)[0]
        
        try:
            while len(d):
                most = d.most_common(1)[0]
                if i<j:
                    for _ in range(most[1]):
                        p[i] = most[0]
                        i += 2
                else:
                    for _ in range(most[1]):
                        p[j] = most[0]
                        j += 2
                d.pop(most[0])
            return ''.join(p)
        except:
            return ''

但是WA，比如每个字母的count为4,4,2，最后会出现2填不进去的情况

错误的原因在于太贪了，每次都把所有的相同字母填进去，note that填进去一个自后，剩余的字母count就要重新计算了

所以就维持一个priority queue实时选取就好了，每次可以实时选取2个字母，也可以选取1个字母（这里以2个为例）

选取1个字母就要注意前后2次选取的字母不要重复

from collections import Counter
import heapq

class Solution:
    def reorganizeString(self, S):
        """
        :type S: str
        :rtype: str
        """
        ret = ''
        pq = []
        c = Counter(S)
        for k, v in c.items():
            heapq.heappush(pq, (-v, k))
        
        prev_key = ''
        while pq:
            v, k = heapq.heappop(pq)
            if k == prev_key:
                if not pq:  return ''
                vv, kk = heapq.heappop(pq)
                heapq.heappush(pq, (v,k))
                v, k = vv, kk
                
            ret += k
            if v+1<0:
                heapq.heappush(pq, (v+1, k))
            prev_key = k
            
        return ret

或者写得更简单点

from collections import Counter
import heapq

class Solution:
    def reorganizeString(self, S):
        """
        :type S: str
        :rtype: str
        """
        ret = ''
        pq = []
        c = Counter(S)
        for k, v in c.items():
            heapq.heappush(pq, (-v, k))
        
        # previous key value
        prev_v, prev_k = 0, ''
        while pq:
            v, k = heapq.heappop(pq)
            ret += k
            if prev_v<0:
                heapq.heappush(pq, (prev_v, prev_k)) # delay one loop to prevent aab
            prev_v, prev_k = v+1, k    # store current key value for latter enqueue
            
        return ret if len(ret)==len(S) else ''