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 = 0, right = 0, dupCount = 0;
        
        while (right < S.size()) {
            if (hash[S.at(right)] > 0) // 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 (1 == 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)] > 1)
                    -- 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 = {4, 4, 2, 1, 2, 3};
    vector<string>  result;
    
    /*
     {abbc }
     {awag naga agaw gwkw wkwa }
     {cc }
     {}
     {aa }
     {aba bab }
     */
    for (auto i = 0; i < sInputs.size(); ++ i) {
        result = testSolution.subStringK1Dist(sInputs[i], iInputs[i]);
        
        cout << "{";
        for (auto it : result)
            cout << it << " ";
        cout << "}" << endl;
    }
    
    return 0;
}