脚本宝典收集整理的这篇文章主要介绍了[LeetCode] 767. Reorganize String，脚本宝典觉得挺不错的，现在分享给大家，也给大家做个参考。
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].
Solution
class Solution {
public String reorganizeString(String S) {
int n = S.length();
int[] cnt = new int[128];
char mc = 'a';
for (char c : S.toCharArray()) {
cnt[c]++;
mc = (cnt[c] > cnt[mc]) ? c : mc;
}
if (cnt[mc] == 1) {
return S;
}
if (cnt[mc] > (n+1)/2) {
return "";
}
StringBuilder[] sb = new StringBuilder[cnt[mc]];
for (int i = 0; i < sb.length; i ++) {
sb[i] = new StringBuilder();
sb[i].append(mc);
}
int k = 0;
for (char c = 'a'; c <= 'z'; c++) {
while (c != mc && cnt[c] > 0) {
sb[k++].append(c);
cnt[c]--;
k %= sb.length;
}
}
for (int i = 1; i < sb.length; i++) {
sb[0].append(sb[i]);
}
return sb[0].toString();
}
}以上是脚本宝典为你收集整理的[LeetCode] 767. Reorganize String全部内容，希望文章能够帮你解决[LeetCode] 767. Reorganize String所遇到的问题。
本图文内容来源于网友网络收集整理提供，作为学习参考使用，版权属于原作者。
如您有任何意见或建议可联系处理。小编QQ：384754419，请注明来意。