C 解法, 执行用时: 4ms, 内存消耗: 408KB, 提交时间: 2022-07-30

#include <stdio.h>

int main() {
    char s[100001] = "";
    scanf("%s", s);
    int h = strlen(s);
    int H = 0;
    int v[26] = { 0 }, c = 0, a = 0, k = 0;
    
    for (int i = 0; i < h; i++) {
        k = s[i] - 'a';
        if (v[k] == 0) {
            v[k] = 1;
            c++;
            if (c == 3) {
                if (i - a > H) {
                    H = i - a;
                }
                while (c == 3) {
                    k = s[a++] - 'a';
                    if (v[k] == 1) {
                        v[k] = 0;
                        c--;
                    } else {
                        v[k] -= 1;
                    }
                }
            }
        } else {
            v[k] += 1;
        }
    }
    
    if (c <= 2) {
        if (h - a > H) {
            H = h - a;
        }
    }
    printf("%d", H);
    return 0;
}

C 解法, 执行用时: 4ms, 内存消耗: 456KB, 提交时间: 2022-06-28

#include <stdio.h>

int main() {
    char s[100001] = "";
    scanf("%s", s);
    
    int sLen = strlen(s);
    int maxLen = 0;
    int hash[26] = { 0 }, chCount = 0, begin = 0, vKey = 0;
    for (int i = 0; i < sLen; i++) {
        vKey = s[i] - 'a';
        if (hash[vKey] == 0) {
            hash[vKey] = 1;
            chCount++;
            if (chCount == 3) {
                if (i - begin > maxLen) {
                    maxLen = i - begin;
                }
                
                while (chCount == 3) {
                    vKey = s[begin++] - 'a';
                    if (hash[vKey] == 1) {
                        hash[vKey] = 0;
                        chCount--;
                    } else {
                        hash[vKey] -= 1;
                    }
                }
            }
        } else {
            hash[vKey] += 1;
        }
    }
    
    if (chCount <= 2) {
        if (sLen - begin > maxLen) {
            maxLen = sLen - begin;
        }
    }
    
    printf("%d", maxLen);
    
    return 0;
}

C 解法, 执行用时: 4ms, 内存消耗: 472KB, 提交时间: 2022-05-03

#include <stdio.h>

//159. 至多包含两个不同字符的最长子串
int lengthOfLongestSubstringTwoDistinct(char* s) {
    int len = strlen(s);
    int maxLen = 0;
    int count = 0;//不同字符的个数
    int hash[128] = {0};
    int left = 0, right = 0;
    while (right < len) {
        hash[s[right]]++;
        if (hash[s[right]] == 1) {
            count++;
        }
        while (count > 2) {
            hash[s[left]]--;
            if (hash[s[left]] == 0) {
                count--;
            }
            left++;
        }
        maxLen = fmax(maxLen, right - left + 1);
        right++;
    }
    return maxLen;
}

int main() {
    int s[100000];
    scanf("%s", &s);
    printf("%d", lengthOfLongestSubstringTwoDistinct(s));
    return 0;
}

C++ 解法, 执行用时: 6ms, 内存消耗: 556KB, 提交时间: 2022-07-29

#include<iostream>
#include<string>
#include<vector>
using namespace std;

int main() {
    string s;
    cin >> s;
    int L = 0, R = 0, c = 0, z = 0;
    vector<int> freq(26);
    
    while (R < s.size()) {
        if (freq[s[R] - 'a'] == 0) {
            c++;
        }
        freq[s[R] - 'a'] ++;
        R++;
        while (c > 2) {
            freq[s[L] - 'a'] --;
            if (freq[s[L] - 'a'] == 0) {
                c --;
            }
            L++;
        }
        z = max(z, R - L);
    }
    cout << z << endl;
}

C++ 解法, 执行用时: 6ms, 内存消耗: 656KB, 提交时间: 2022-07-12

#include<iostream>
#include<string>
#include<vector>
using namespace std;

int main(){
    string s;
    cin >> s;
    int l = 0, r = 0, count = 0, res = 0;
    vector<int> freq(26);
    while(r < s.size()){
        if(freq[s[r] - 'a'] == 0){
            count ++;
        }
        freq[s[r] - 'a'] ++;
        r ++;
        while(count > 2){
            freq[s[l] - 'a'] --;
            if(freq[s[l] - 'a'] == 0){
                count --;
            }
            l ++;
        }
        res = max(res, r - l);
    }
//     return res;
    cout << res << endl;
}