767. Reorganize String

https://leetcode.com/problems/reorganize-string/description/
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].

---

这题有两种方法，用PriorityQueue比较容易理解。

approach1: priorityqueue

    //#2 using PriorityQueue, tracking the remaining 2 most frequent numbers
    public String reorganizeString(String S) {
        if (S == null || S.length() <= 1) {
            return S;
        }
        int len = S.length();
        //attention: array size is 26! not len
        int[] counts = new int[26];
        for (int i = 0; i < S.length(); i++) {
            counts[S.charAt(i) - 'a']++;
            //attention 2: exit condition
            if (counts[S.charAt(i) - 'a'] > (len + 1) / 2) return "";
        }

        PriorityQueue priorityQueue = new PriorityQueue<>(len, new Comparator() {
            @Override
            public int compare(RichChar o1, RichChar o2) {
                return o2.count - o1.count;
            }
        });
        for (int i = 0; i < 26; i++) {
            if (counts[i] != 0) {
                priorityQueue.offer(new RichChar((char) (i + 'a'), counts[i]));
            }
        }

        //attention, a stringbuilder is used here
        StringBuilder res = new StringBuilder();
        while (priorityQueue.size() >= 2) {
            RichChar ch1 = priorityQueue.poll();
            RichChar ch2 = priorityQueue.poll();
            res.append(ch1.letter);
            res.append(ch2.letter);
            ch1.count--;
            ch2.count--;
            if (ch1.count > 0) priorityQueue.offer(ch1);
            if (ch2.count > 0) priorityQueue.offer(ch2);
        }
        if (priorityQueue.size() > 0) {
            res.append(priorityQueue.poll().letter);
        }
        return res.toString();
    }

    class RichChar {
        char letter;
        int count;

        RichChar(char letter, int count) {
            this.letter = letter;
            this.count = count;
        }
    }

approach 2: 插值

隔一个slot插值：

    //#1
    public String reorganizeString0(String S) {
        int N = S.length();
        int[] counts = new int[26];
        //aabbc -> [200, 200, 100, 0, 0..]，保证个数多的在前面
        for (char c : S.toCharArray()) {
            counts[c - 'a'] += 100;
        }
        //aabbc -> [200, 201, 102, 3, 4..] 个数相同的情况下按照alphabet排列
        for (int i = 0; i < 26; ++i) {
            counts[i] += i;
        }
        //Encoded counts[i] = 100*(actual count) + (i)
        Arrays.sort(counts);

        char[] ans = new char[N];
        int t = 1;
        for (int code : counts) {
            int ct = code / 100;
            char ch = (char) ('a' + (code % 100));
            if (ct > (N + 1) / 2) return "";
            for (int i = 0; i < ct; ++i) {
                if (t >= N)
                    t = 0;
                ans[t] = ch;
                t += 2;
            }
        }

        return String.valueOf(ans);
    }

具体可以看下这题的solution。