You are given two strings s and t of the same length. You want to change s to t. Changing the i-th character of s to i-th character of t costs |s[i] - t[i]| that is, the absolute difference between the ASCII values of the characters.
You are also given an integer maxCost.
Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of twith a cost less than or equal to maxCost.
If there is no substring from s that can be changed to its corresponding substring from t, return 0.
Example 1:
Input: s = "abcd", t = "bcdf", maxCost = 3 Output: 3 Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.
Example 2:
Input: s = "abcd", t = "cdef", maxCost = 3
Output: 1
Explanation: Each character in s costs 2 to change to charactor in t, so the maximum length is 1.
Example 3:
Input: s = "abcd", t = "acde", maxCost = 0 Output: 1 Explanation: You can't make any change, so the maximum length is 1.
Constraints:
1 <= s.length, t.length <= 10^50 <= maxCost <= 10^6-
sandtonly contain lower case English letters.
尽可能使字符串相等。
给你两个长度相同的字符串，s 和 t。
将 s 中的第 i 个字符变到 t 中的第 i 个字符需要 |s[i] - t[i]| 的开销（开销可能为 0），也就是两个字符的 ASCII 码值的差的绝对值。
用于变更字符串的最大预算是 maxCost。在转化字符串时，总开销应当小于等于该预算，这也意味着字符串的转化可能是不完全的。
如果你可以将 s 的子字符串转化为它在 t 中对应的子字符串，则返回可以转化的最大长度。
如果 s 中没有子字符串可以转化成 t 中对应的子字符串，则返回 0。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/get-equal-substrings-within-budget
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
思路是滑动窗口，而且可以用模板。模板参见76题。可以消耗的开销一共是maxCost，对于每个相同位置上的字符的ASCII 码的差值，可以通过先移动 end 指针不断消耗 maxCost，直到maxCost == 0。当maxCost == 0的时候，可以试图移动 start 指针，过程中记录 res 的最大值即可。
时间O(n)
空间O(1)
Java实现
1 class Solution {
2 public int equalSubstring(String s, String t, int maxCost) {
3 int len = s.length();
4 int res = 0;
5 int start = 0;
6 int end = 0;
7 while (end < len) {
8 maxCost -= Math.abs(s.charAt(end) - t.charAt(end));
9 end++;
10 while (maxCost < 0) {
11 maxCost += Math.abs(s.charAt(start) - t.charAt(start));
12 start++;
13 }
14 res = Math.max(res, end - start);
15 }
16 return res;
17 }
18 }