Find substring with K-1 distinct characters

参考 Find substring with K distinct characters

Find substring with K distinct characters(http://www.cnblogs.com/pegasus923/p/8444653.html)

Given a string and number K, find the substrings of size K with K-1 distinct characters. If no, output empty list. Remember to emit the duplicate substrings, i.e. if the substring repeated twice, only output once.

• 字符串中等题。Sliding window algorithm + Hash。此题跟上题的区别在于，子串中有一个重复字符。
• 思路还是跟上题一样，只是需要把对count的判断条件改成dupCount。当窗口size为K时，如果重复字符只有一个的话，则为结果集。对dupCount操作的判断条件，也需要改为>0, >1。

 //
 //  main.cpp
 //  LeetCode
 //
 //  Created by Hao on 2017/3/16.
 //  Copyright © 2017年 Hao. All rights reserved.
 //
 
 #include <iostream>
 #include <vector>
 #include <unordered_map>
 using namespace std;
 
 class Solution {
 public:
     vector<string> subStringK1Dist(string S, int K) {
         vector<string> vResult;
 
         // corner case
         if (S.empty()) return vResult;
 
         unordered_map<char, int>    hash;
 
         // window start/end pointer, hit count
         int left = , right = , dupCount = ;
 
         while (right < S.size()) {
             if (hash[S.at(right)] > ) // hit the condition dup char
                 ++ dupCount;
 
             ++ hash[S.at(right)]; // count the occurrence
 
             ++ right; // move window end pointer rightward
 
             // window size reaches K
             if (right - left == K) {
                 if ( == dupCount) { // find 1 dup char
                     if (find(vResult.begin(), vResult.end(), S.substr(left, K)) == vResult.end()) // using STL find() to avoid dup
                         vResult.push_back(S.substr(left, K));
                 }
 
                 // dupCount is only increased when hash[i] > 0, so only hash[i] > 1 means that dupCount was increased.
                 if (hash[S.at(left)] > )
                     -- dupCount;
 
                 -- hash[S.at(left)]; // decrease to restore occurrence
 
                 ++ left; // move window start pointer rightward
             }
         }
 
         return vResult;
     }
 };
 
 int main(int argc, char* argv[])
 {
     Solution    testSolution;
 
     vector<string>  sInputs = {"aaabbcccc","awaglknagawunagwkwagl", "abccdef", "", "aaaaaaa", "ababab"};
     vector<int>     iInputs = {, , , , , };
     vector<string>  result;
 
     /*
      {abbc }
      {awag naga agaw gwkw wkwa }
      {cc }
      {}
      {aa }
      {aba bab }
      */
     for (auto i = ; i < sInputs.size(); ++ i) {
         result = testSolution.subStringK1Dist(sInputs[i], iInputs[i]);
 
         cout << "{";
         for (auto it : result)
             cout << it << " ";
         cout << "}" << endl;
     }
 
     return ;
 }