【leetcode】424. Longest Repeating Character Replacement

题目如下：

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.

解题思路：对于任意一个区间[i,j]能否经过至多K次替换可以变成只包含同一个字符的子串，只要判断除去其中某一个字符，其余剩下的字符的总和不超过K即可。所以本题的解题就是就是找到符合这个条件的最长区间。怎么找这个区间？我的方法是引入low和high两个index，从字符串头部开始遍历，依次判断[low,high]区间是否满足条件，满足的话high += 1，不满足则low += 1。

代码如下：

class Solution(object):
    def check(self,dic,total,k):
        mv = 0
        flag = False
        for i in dic.iterkeys():
            if total - dic[i] <= k:
                flag = True
                mv = max(mv,total)
        if flag == False and len(dic) > 0:
            return -1
        return mv

    def characterReplacement(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: int
        """
        if len(s) == 0:
            return 0
        low = 0
        high = 1
        dic = {}
        dic[s[0]] = 1
        total = 1
        res = 0
        s += '#'
        while low < high and high < len(s):
            rc = self.check(dic,total,k)
            if rc == -1:
                dic[s[low]] -= 1
                total -= 1
                if dic[s[low]] == 0:
                    del dic[s[low]]
                low += 1
            else:
                dic[s[high]] = dic.setdefault(s[high],0) + 1
                total += 1
                high += 1

            res = max(res,rc)
        return res