leetcode_[python/C++]_424_Longest Repeating Character Replacement

题目链接
【题目】
Given a string that consists of only uppercase English letters, you can replace any letter in the string with another letter at most k times. Find the length of a longest substring containing all repeating letters you can get after performing the above operations.

---

Note:
Both the string’s length and k will not exceed 104.

Example 1:

Input:
s = “ABAB”, k = 2

Output:
4

Explanation:
Replace the two ‘A’s with two ‘B’s or vice versa.

---

Example 2:

Input:
s = “AABABBA”, k = 1

Output:
4

Explanation:
Replace the one ‘A’ in the middle with ‘B’ and form “AABBBBA”.
The substring “BBBB” has the longest repeating letters, which is 4.
Subscribe to see which companies asked this question

---

【分析】
这道题主要是为了分享一种很不错的解法，叫做滑窗算法，类似我们计算机网络中的发送数据包的滑动窗口移动。
算法思想：
即定义end和begin指针，end -begin即为窗体的长度，始终维护一个窗体，要保证该窗体中：窗体长度 - 最多字母个数 > k，也就是维护窗体中除了个数最多的字符外，其他字符总数不超过k。窗体长度即为最多变换k次之后，最长的相同字母字符串。

class Solution {public:    int characterReplacement(string s, int k) {        int size = s.size();        vector<int> count(26,0);        int begin = 0, maxlen = 0, ans = 0;        for(int end=0;end<size;end++){            maxlen = max(maxlen,++count[s[end]-'A']);            while(end - begin + 1 - maxlen > k) count[s[begin++] - 'A']--;//相当于count[s[begin] - 'A']--;  begin ++;            ans = max(ans,  end - begin + 1);        }        return ans;    }};

---

python写法

class Solution(object):    def characterReplacement(self,s, k):        """        :type s: str        :type k: int        :rtype: int        """        from math import *        size = len(s)        count = [0]*26        begin = 0        maxlen = 0        ans = 0        for end in range(size):            count[ord(s[end])-ord('A')] = count[ord(s[end])-ord('A')] + 1            maxlen = max(maxlen,count[ord(s[end])-ord('A')])            while end - begin + 1 - maxlen > k:                count[ord(s[begin])-ord('A')] = count[ord(s[begin])-ord('A')] - 1                begin = begin + 1            ans = max(ans,end - begin + 1)        return ans

discuss中有一种利用collections.counters()的做法，很强

class Solution(object):    def characterReplacement(self, s, k):        res = lo = 0        counts = collections.Counter()        for hi in range(len(s)):            counts[s[hi]] += 1            max_char_n = counts.most_common(1)[0][1]            while (hi - lo - max_char_n + 1 > k):                counts[s[lo]] -= 1                lo += 1            res = max(res, hi - lo + 1)        return res

这道题一开始看错题目了，以为k是指能改变一种字母多少次，就写了下面这种代码，有点粗糙但是符合这种要求的解法

int characterReplacement(string s, int k) {        set<char> sets;        int size = s.size();        for(int i=0;i<size;i++){            sets.insert(s[i]);        }        char chr;        set<char>::iterator iter;        char ans_chr;        int begin_index = 0;        int end_index = 0;        int max = -100000;        for(iter = sets.begin();iter!= sets.end();iter++){            chr = *iter;            vector<int> dp(size,0);            vector<int> first(sets.size(),0);            int num[124];            memset(num,0,sizeof(num));            for(int i=1;i<size;i++){                if(num[s[i]] == 0) first[s[i]] = i;                if(s[i]==chr){                    dp[i] = dp[i-1];                    if(i==1 && s[i]!=s[i-1]) dp[i] = 1;                }                else{                    if(num[s[i]] < k){                        num[s[i]]++;                        dp[i] = dp[i-1];                    }                    else{                        if(first[s[i]] == 0){                            dp[i] = i;                            memset(num,0,sizeof(num));                            first[s[i]] = i;                        }                        else{                            dp[i] = first[s[i]]+1;                            memset(num,0,sizeof(num));                            for(int j = first[s[i]]+1;j<=i;j++){                                if(num[s[j]] == 0){                                    first[s[j]] = j;                                }                                num[s[j]]++;                            }                        }                    }                }            }            for(int i=0;i<size;i++){                if(i-dp[i]>max){                    begin_index = dp[i];                    end_index = i;                    max = i - dp[i];                    ans_chr = chr;                }            }        }        return max+1;}