From 01e0740a7b23309c606c2e9bc2fec7d5e8a15cad Mon Sep 17 00:00:00 2001 From: YDZ Date: Fri, 25 Oct 2019 11:55:13 +0800 Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20problem=201111?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ... Depth of Two Valid Parentheses Strings.go | 49 ++++++++++++ ...h of Two Valid Parentheses Strings_test.go | 47 ++++++++++++ .../README.md | 75 +++++++++++++++++++ 3 files changed, 171 insertions(+) create mode 100644 Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go create mode 100644 Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go create mode 100755 Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/README.md diff --git a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go b/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go new file mode 100644 index 00000000..6268f186 --- /dev/null +++ b/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go @@ -0,0 +1,49 @@ +package leetcode + +// 解法一 二分思想 +func maxDepthAfterSplit(seq string) []int { + stack, maxDepth, res := 0, 0, []int{} + for _, v := range seq { + if v == '(' { + stack++ + maxDepth = max(stack, maxDepth) + } else { + stack-- + } + } + stack = 0 + for i := 0; i < len(seq); i++ { + if seq[i] == '(' { + stack++ + if stack <= maxDepth/2 { + res = append(res, 0) + } else { + res = append(res, 1) + } + } else { + if stack <= maxDepth/2 { + res = append(res, 0) + } else { + res = append(res, 1) + } + stack-- + } + } + return res +} + +// 解法二 模拟 +func maxDepthAfterSplit1(seq string) []int { + stack, top, res := make([]int, len(seq)), -1, make([]int, len(seq)) + for i, r := range seq { + if r == ')' { + res[i] = res[stack[top]] + top-- + continue + } + top++ + stack[top] = i + res[i] = top % 2 + } + return res +} diff --git a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go b/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go new file mode 100644 index 00000000..fffc33e8 --- /dev/null +++ b/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go @@ -0,0 +1,47 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question1111 struct { + para1111 + ans1111 +} + +// para 是参数 +// one 代表第一个参数 +type para1111 struct { + one string +} + +// ans 是答案 +// one 代表第一个答案 +type ans1111 struct { + one []int +} + +func Test_Problem1111(t *testing.T) { + + qs := []question1111{ + + question1111{ + para1111{"(()())"}, + ans1111{[]int{0, 1, 1, 1, 1, 0}}, + }, + + question1111{ + para1111{"()(())()"}, + ans1111{[]int{0, 0, 0, 1, 1, 0, 1, 1}}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 1111------------------------\n") + + for _, q := range qs { + _, p := q.ans1111, q.para1111 + fmt.Printf("【input】:%v 【output】:%v\n", p, maxDepthAfterSplit(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/README.md b/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/README.md new file mode 100755 index 00000000..56bdf8d6 --- /dev/null +++ b/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/README.md @@ -0,0 +1,75 @@ +# [1111. Maximum Nesting Depth of Two Valid Parentheses Strings](https://leetcode.com/problems/maximum-nesting-depth-of-two-valid-parentheses-strings/) + + +## 题目: + +A string is a *valid parentheses string* (denoted VPS) if and only if it consists of `"("` and `")"` characters only, and: + +- It is the empty string, or +- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are VPS's, or +- It can be written as `(A)`, where `A` is a VPS. + +We can similarly define the *nesting depth* `depth(S)` of any VPS `S` as follows: + +- `depth("") = 0` +- `depth(A + B) = max(depth(A), depth(B))`, where `A` and `B` are VPS's +- `depth("(" + A + ")") = 1 + depth(A)`, where `A` is a VPS. + +For example, `""`, `"()()"`, and `"()(()())"` are VPS's (with nesting depths 0, 1, and 2), and `")("` and `"(()"` are not VPS's. + +Given a VPS seq, split it into two disjoint subsequences `A` and `B`, such that `A` and `B` are VPS's (and `A.length + B.length = seq.length`). + +Now choose **any** such `A` and `B` such that `max(depth(A), depth(B))` is the minimum possible value. + +Return an `answer` array (of length `seq.length`) that encodes such a choice of `A` and `B`: `answer[i] = 0` if `seq[i]` is part of `A`, else `answer[i] = 1`. Note that even though multiple answers may exist, you may return any of them. + +**Example 1:** + + Input: seq = "(()())" + Output: [0,1,1,1,1,0] + +**Example 2:** + + Input: seq = "()(())()" + Output: [0,0,0,1,1,0,1,1] + +**Constraints:** + +- `1 <= seq.size <= 10000` + + +## 题目大意 + + +有效括号字符串 仅由 "(" 和 ")" 构成，并符合下述几个条件之一： + +- 空字符串 +- 连接，可以记作 AB（A 与 B 连接），其中 A 和 B 都是有效括号字符串 +- 嵌套，可以记作 (A)，其中 A 是有效括号字符串 + +类似地，我们可以定义任意有效括号字符串 s 的 嵌套深度 depth(S)： + +- s 为空时，depth("") = 0 +- s 为 A 与 B 连接时，depth(A + B) = max(depth(A), depth(B))，其中 A 和 B 都是有效括号字符串 +- s 为嵌套情况，depth("(" + A + ")") = 1 + depth(A)，其中 A 是有效括号字符串 + + +例如：""，"()()"，和 "()(()())" 都是有效括号字符串，嵌套深度分别为 0，1，2，而 ")(" 和 "(()" 都不是有效括号字符串。 + +  + +给你一个有效括号字符串 seq，将其分成两个不相交的子序列 A 和 B，且 A 和 B 满足有效括号字符串的定义（注意：A.length + B.length = seq.length）。 + +现在，你需要从中选出 任意 一组有效括号字符串 A 和 B，使 max(depth(A), depth(B)) 的可能取值最小。 + +返回长度为 seq.length 答案数组 answer ，选择 A 还是 B 的编码规则是：如果 seq[i] 是 A 的一部分，那么 answer[i] = 0。否则，answer[i] = 1。即便有多个满足要求的答案存在，你也只需返回 一个。 + + + +## 解题思路 + +- 给出一个括号字符串。选出 A 部分和 B 部分，使得 `max(depth(A), depth(B))` 值最小。在最终的数组中输出 0 和 1，0 标识是 A 部分，1 标识是 B 部分。 +- 这一题想要 `max(depth(A), depth(B))` 值最小，可以使用贪心思想。如果 A 部分和 B 部分都尽快括号匹配，不深层次嵌套，那么总的层次就会变小。只要让嵌套的括号中属于 A 的和属于 B 的间隔排列即可。例如：“`(((())))`”，上面的字符串的嵌套深度是 4，按照上述的贪心思想，则标记为 0101 1010。 +- 这一题也可以用二分的思想来解答。把深度平分给 A 部分和 B 部分。 + - 第一次遍历，先计算最大深度 + - 第二次遍历，把深度小于等于最大深度一半的括号标记为 0(给 A 部分)，否则标记为 1(给 B 部分)