Given a string and a positive integer d. Some characters may be repeated in the given string. Rearrange characters of the given string such that the same characters become d distance away from each other. Note that there can be many possible rearrangements, the output should be one of the possible rearrangements. If no such arrangement is possible, that should also be reported.
Expected time complexity is O(n) where n is length of input string.

Examples:
Input:? "abb", d = 2
Output: "bab"

Input:? "aacbbc", d = 3
Output: "abcabc"

Input: "geeksforgeeks", d = 3
Output: egkegkesfesor

Input:? "aaa",? d = 2
Output: Cannot be rearranged
摘自：http://www.geeksforgeeks.org/rearrange-a-string-so-that-all-same-characters-become-at-least-d-distance-away/

這題就是貪心算法。對(duì)于字符串處理來(lái)講，如果不care最終的符號(hào)序，就要以考慮用統(tǒng)計(jì)重排的方法。

1. 統(tǒng)計(jì)所有字符的出現(xiàn)頻率；

2. 按出現(xiàn)頻率從高到低優(yōu)先填好所有字符；

如果全部都是ASCII碼，共出現(xiàn)m個(gè)不同字符，第2步，可以：

1. 用最大堆來(lái)做，也可以直接排序，時(shí)間復(fù)雜度都是O(n+mlgm)，空間復(fù)雜度就是O(m)。因?yàn)閙<256<<n，所以最終復(fù)雜度也可以說(shuō)是O(n)。

2. 也可以直接用一個(gè)vector<list<char> >記錄每個(gè)頻率下的字符來(lái)做。用vector<list<char> >就是O(2n)的時(shí)間復(fù)雜度了。空間復(fù)雜度高點(diǎn)，O(n)。

?

Rearrange a string so that all same characters become d distance away