添加 problem 1208

Algorithms/1208. Get Equal Substrings Within Budget/1208. Get Equal Substrings Within Budget.go

@@ -0,0 +1,16 @@
+package leetcode
+
+func equalSubstring(s string, t string, maxCost int) int {
+	left, right, res := 0, -1, 0
+	for left < len(s) {
+		if right+1 < len(s) && maxCost-abs(int(s[right+1]-'a')-int(t[right+1]-'a')) >= 0 {
+			right++
+			maxCost -= abs(int(s[right]-'a') - int(t[right]-'a'))
+		} else {
+			res = max(res, right-left+1)
+			maxCost += abs(int(s[left]-'a') - int(t[left]-'a'))
+			left++
+		}
+	}
+	return res
+}

Algorithms/1208. Get Equal Substrings Within Budget/1208. Get Equal Substrings Within Budget_test.go

@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1208 struct {
+	para1208
+	ans1208
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1208 struct {
+	s       string
+	t       string
+	maxCost int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1208 struct {
+	one int
+}
+
+func Test_Problem1208(t *testing.T) {
+
+	qs := []question1208{
+
+		question1208{
+			para1208{"abcd", "bcdf", 3},
+			ans1208{3},
+		},
+
+		question1208{
+			para1208{"abcd", "cdef", 3},
+			ans1208{1},
+		},
+
+		question1208{
+			para1208{"abcd", "acde", 0},
+			ans1208{1},
+		},
+
+		question1208{
+			para1208{"thjdoffka", "qhrnlntls", 11},
+			ans1208{3},
+		},
+
+		question1208{
+			para1208{"krrgw", "zjxss", 19},
+			ans1208{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1208------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1208, q.para1208
+		fmt.Printf("【input】:%v       【output】:%v\n", p, equalSubstring(p.s, p.t, p.maxCost))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/1208. Get Equal Substrings Within Budget/README.md

@@ -0,0 +1,53 @@
+# [1208. Get Equal Substrings Within Budget](https://leetcode.com/problems/get-equal-substrings-within-budget/)
+
+
+## 题目:
+
+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 `t`with 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 <= 10^5`
+- `0 <= maxCost <= 10^6`
+- `s` and `t` only contain lower case English letters.
+
+## 题目大意
+
+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。
+
+提示：
+
+- 1 <= s.length, t.length <= 10^5
+- 0 <= maxCost <= 10^6
+- s 和 t 都只含小写英文字母。
+
+## 解题思路
+
+- 给出 2 个字符串 `s` 和 `t` 和一个“预算”，要求把“预算”尽可能的花完，`s` 中最多连续有几个字母能变成 `t` 中的字母。“预算”的定义是：|s[i] - t[i]| 。
+- 这一题是滑动窗口的题目，滑动窗口右边界每移动一格，就减少一定的预算，直到预算不能减少，再移动滑动窗口的左边界，这个时候注意要把预算还原回去。当整个窗口把字符 `s` 或 `t` 都滑动完了的时候，取出滑动过程中窗口的最大值即为结果。