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 = "aabbcc", 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.
Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.
这道题给了我们一个字符串str，和一个整数k，让我们对字符串str重新排序，使得其中相同的字符之间的距离不小于k，这道题的难度标为Hard，看来不是省油的灯。的确，这道题的解法用到了哈希表，堆，和贪婪算法。这道题我最开始想的算法没有通过OJ的大集合超时了，下面的方法是参考网上大神的解法，发现十分的巧妙。我们需要一个哈希表来建立字符和其出现次数之间的映射，然后需要一个堆来保存这每一堆映射，按照出现次数来排序。然后如果堆不为空我们就开始循环，我们找出k和str长度之间的较小值，然后从0遍历到这个较小值，对于每个遍历到的值，如果此时堆为空了，说明此位置没法填入字符了，返回空字符串，否则我们从堆顶取出一对映射，然后把字母加入结果res中，此时映射的个数减1，如果减1后的个数仍大于0，则我们将此映射加入临时集合v中，同时str的个数len减1，遍历完一次，我们把临时集合中的映射对由加入堆中，参见代码如下：
class Solution {
public:
string rearrangeString(string str, int k) {
if (k == 0) return str;
string res;
int len = (int)str.size();
unordered_map<char, int> m;
priority_queue<pair<int, char>> q;
for (auto a : str) ++m[a];
for (auto it = m.begin(); it != m.end(); ++it) {
q.push({it->second, it->first});
}
while (!q.empty()) {
vector<pair<int, int>> v;
int cnt = min(k, len);
for (int i = 0; i < cnt; ++i) {
if (q.empty()) return "";
auto t = q.top(); q.pop();
res.push_back(t.second);
if (--t.first > 0) v.push_back(t);
--len;
}
for (auto a : v) q.push(a);
}
return res;
}
};
参考资料：
https://leetcode.com/discuss/108174/c-unordered_map-priority_queue-solution-using-cache
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