358. Rearrange String k Distance Apart

题目：
Given a non-empty string str and an integer k, rearrange the string such that the same characters are at least distance k from each other.

All input strings are given in lowercase letters. If it is not possible to rearrange the string, return an empty string "".

Example 1:
str = " ", k = 3

Result: "abcabc"

The same letters are at least distance 3 from each other.
Example 2:
str = "aaabc", k = 3

Answer: ""

It is not possible to rearrange the string.
Example 3:
str = "aaadbbcc", k = 2

Answer: "abacabcd"

Another possible answer is: "abcabcda"

The same letters are at least distance 2 from each other.

解答：

//先记录str中的char及它出现在次数，存在count[]里，用valid[]来记录这个char最小出现的位置。
    //每一次把count值最大的数选出来，append到新的string后面
    public int selectedValue(int[] count, int[] valid, int i) {
        int select = Integer.MIN_VALUE;
        int val = -1;
        for (int j = 0; j < count.length; j++) {
            if (count[j] > 0 && i >= valid[j] && count[j] > select) {
                select = count[j];
                val = j;
            }
        }
        return val;
    }
    
    public String rearrangeString(String str, int k) {
        int[] count = new int[26];
        int[] valid = new int[26];
        //把每个出现了的char的个数记下来
        for (char c : str.toCharArray()) {
            count[c - 'a']++;
        }
        
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < str.length(); i++) {
            //选出剩下需要出现次数最多又满足条件的字母，即是我们最应该先放的数
            int curt = selectedValue(count, valid, i);
            //如果不符合条件，返回“”
            if (curt == -1) return "";
            //选择好后，count要减少，valid要到下一个k distance之后
            count[curt]--;
            valid[curt] = i + k;
            sb.append((char)('a' + curt));
        }
        
        return sb.toString();
    }