LWC 68: 767. Reorganize String

LWC 68: 767. Reorganize String

传送门：767. Reorganize String

Problem:

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：
这是一种选择策略，如果统计字符出现的次数发现，约束出现频次最高的字符在tuple<a, b>中的位置a时，可以保证后续tuple中，位置a上的字符永远不会和前一个tuple中，位置b上的字符出现重复。按照这种约束关系选择到最后，如果还剩下一个字符的频次大于1的情况，说明无法构成Reorganize String。

代码如下：

class P implements Comparable<P>{
        int c;
        int count;
        P(int c, int count){
            this.c = c;
            this.count = count;
        }
        @Override
        public int compareTo(P o) {
            return o.count - this.count;
        }

        @Override
        public String toString() {
            return c + "," + count;
        }
    }

    public String reorganizeString(String S) {
        P[] ps = new P[32];
        for (int i = 0; i < 32; ++i) ps[i] = new P(i, 0);
        for (char c : S.toCharArray()) {
            int key = c - 'a';
            ps[key].count ++;
        }
        PriorityQueue<P> queue = new PriorityQueue<>();
        for (int i = 0; i < 32; ++i) {
            if (ps[i].count >= 1)
                queue.offer(ps[i]);
        }   
        StringBuilder sb = new StringBuilder();
        while (!queue.isEmpty()) {
            P fir = queue.poll(); // fir
            if (queue.isEmpty()) {
                if (fir.count >= 2) return "";
                sb.append((char)(fir.c + 'a'));
            }
            else {
                P sec = queue.poll(); // sec
                sb.append((char)(fir.c + 'a'));
                sb.append((char)(sec.c + 'a'));
                fir.count --;
                sec.count --;
                if (fir.count >= 1) {
                    queue.offer(fir);
                }
                if (sec.count >= 1) {
                    queue.offer(sec);
                }
            }
        }
        return sb.toString();
    }

Python版本：

class Solution(object):
    def reorganizeString(self, S):
        """
        :type S: str
        :rtype: str
        """
        from collections import Counter
        import heapq

        res = ""
        pq  = []
        c   = Counter(S)
        for key, value in c.items():
            heapq.heappush(pq, (-value, key))
        while pq:
            val_1, key_1 = heapq.heappop(pq)
            val_1 = -val_1
            if len(pq) == 0:
                if val_1 >= 2: return ""
                res += key_1
            else:
                val_2, key_2 = heapq.heappop(pq)
                val_2 = -val_2
                res += key_1
                res += key_2
                val_1 -= 1
                val_2 -= 1
                if val_1 >= 1: heapq.heappush(pq, (-val_1, key_1))
                if val_2 >= 1: heapq.heappush(pq, (-val_2, key_2))
        return res