From 2331702b1574a21e75a90df89d81108f9ebae9e9 Mon Sep 17 00:00:00 2001 From: YDZ Date: Tue, 27 Aug 2019 17:22:33 +0800 Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20problem=20978?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../498. Diagonal Traverse.go | 7 +-- Algorithms/0509. Fibonacci Number/README.md | 6 +- .../978. Longest Turbulent Subarray.go | 43 +++++++++++++ .../978. Longest Turbulent Subarray_test.go | 62 +++++++++++++++++++ .../README.md | 58 +++++++++++++++++ 5 files changed, 166 insertions(+), 10 deletions(-) create mode 100644 Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray.go create mode 100644 Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray_test.go create mode 100755 Algorithms/0978. Longest Turbulent Subarray/README.md diff --git a/Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse.go b/Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse.go index 4af24378..98bae54f 100644 --- a/Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse.go +++ b/Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse.go @@ -5,15 +5,12 @@ func findDiagonalOrder1(matrix [][]int) []int { if matrix == nil || len(matrix) == 0 || len(matrix[0]) == 0 { return nil } - row := len(matrix) - col := len(matrix[0]) - dir := [2][2]int{ + row, col, dir, i, x, y, d := len(matrix), len(matrix[0]), [2][2]int{ {-1, 1}, {1, -1}, - } + }, 0, 0, 0, 0 total := row * col res := make([]int, total) - var i, x, y, d int for i < total { for x >= 0 && x < row && y >= 0 && y < col { res[i] = matrix[x][y] diff --git a/Algorithms/0509. Fibonacci Number/README.md b/Algorithms/0509. Fibonacci Number/README.md index ff0ca5da..4f37321a 100755 --- a/Algorithms/0509. Fibonacci Number/README.md +++ b/Algorithms/0509. Fibonacci Number/README.md @@ -1,9 +1,5 @@ -# 509. Fibonacci Number +# [509. Fibonacci Number](https://leetcode.com/problems/fibonacci-number/) -Created Time: Aug 24, 2019 2:52 PM -Difficulty: Easy -Link: https://leetcode.com/problems/fibonacci-number/ -Tags: Array,Dynamic Programming,Math,Recursive ## 题目: diff --git a/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray.go b/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray.go new file mode 100644 index 00000000..6c04a15c --- /dev/null +++ b/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray.go @@ -0,0 +1,43 @@ +package leetcode + +// 解法一 模拟法 +func maxTurbulenceSize(A []int) int { + inc, dec := 1, 1 + maxLen := min(1, len(A)) + for i := 1; i < len(A); i++ { + if A[i-1] < A[i] { + inc = dec + 1 + dec = 1 + } else if A[i-1] > A[i] { + dec = inc + 1 + inc = 1 + } else { + inc = 1 + dec = 1 + } + maxLen = max(maxLen, max(inc, dec)) + } + return maxLen +} + +// 解法二 滑动窗口 +func maxTurbulenceSize1(A []int) int { + if len(A) == 1 { + return 1 + } + // flag > 0 代表下一个数要大于前一个数，flag < 0 代表下一个数要小于前一个数 + res, left, right, flag, lastNum := 0, 0, 0, A[1]-A[0], A[0] + for left < len(A) { + if right < len(A)-1 && ((A[right+1] > lastNum && flag > 0) || (A[right+1] < lastNum && flag < 0) || (right == left)) { + right++ + flag = lastNum - A[right] + lastNum = A[right] + } else { + if right != left && flag != 0 { + res = max(res, right-left+1) + } + left++ + } + } + return max(res, 1) +} diff --git a/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray_test.go b/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray_test.go new file mode 100644 index 00000000..cd3c1e80 --- /dev/null +++ b/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray_test.go @@ -0,0 +1,62 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question978 struct { + para978 + ans978 +} + +// para 是参数 +// one 代表第一个参数 +type para978 struct { + one []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans978 struct { + one int +} + +func Test_Problem978(t *testing.T) { + + qs := []question978{ + + question978{ + para978{[]int{0, 1, 1, 0, 1, 0, 1, 1, 0, 0}}, + ans978{5}, + }, + + question978{ + para978{[]int{9, 9}}, + ans978{1}, + }, + + question978{ + para978{[]int{9, 4, 2, 10, 7, 8, 8, 1, 9}}, + ans978{5}, + }, + + question978{ + para978{[]int{4, 8, 12, 16}}, + ans978{2}, + }, + + question978{ + para978{[]int{100}}, + ans978{1}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 978------------------------\n") + + for _, q := range qs { + _, p := q.ans978, q.para978 + fmt.Printf("【input】:%v 【output】:%v\n", p, maxTurbulenceSize(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/Algorithms/0978. Longest Turbulent Subarray/README.md b/Algorithms/0978. Longest Turbulent Subarray/README.md new file mode 100755 index 00000000..4b0bbd49 --- /dev/null +++ b/Algorithms/0978. Longest Turbulent Subarray/README.md @@ -0,0 +1,58 @@ +# [978. Longest Turbulent Subarray](https://leetcode.com/problems/longest-turbulent-subarray/) + +## 题目: + +A subarray `A[i], A[i+1], ..., A[j]` of `A` is said to be *turbulent* if and only if: + +- For `i <= k < j`, `A[k] > A[k+1]` when `k` is odd, and `A[k] < A[k+1]` when `k` is even; +- **OR**, for `i <= k < j`, `A[k] > A[k+1]` when `k` is even, and `A[k] < A[k+1]` when `k` is odd. + +That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray. + +Return the **length** of a maximum size turbulent subarray of A. + +**Example 1:** + + Input: [9,4,2,10,7,8,8,1,9] + Output: 5 + Explanation: (A[1] > A[2] < A[3] > A[4] < A[5]) + +**Example 2:** + + Input: [4,8,12,16] + Output: 2 + +**Example 3:** + + Input: [100] + Output: 1 + +**Note:** + +1. `1 <= A.length <= 40000` +2. `0 <= A[i] <= 10^9` + + +## 题目大意 + + +当 A 的子数组 A[i], A[i+1], ..., A[j] 满足下列条件时，我们称其为湍流子数组： + +若 i <= k < j，当 k 为奇数时， A[k] > A[k+1]，且当 k 为偶数时，A[k] < A[k+1]； +或 若 i <= k < j，当 k 为偶数时，A[k] > A[k+1] ，且当 k 为奇数时， A[k] < A[k+1]。 +也就是说，如果比较符号在子数组中的每个相邻元素对之间翻转，则该子数组是湍流子数组。 + +返回 A 的最大湍流子数组的长度。 + +提示： + +- 1 <= A.length <= 40000 +- 0 <= A[i] <= 10^9 + + + +## 解题思路 + + +- 给出一个数组，要求找出“摆动数组”的最大长度。所谓“摆动数组”的意思是，元素一大一小间隔的。 +- 这一题可以用滑动窗口来解答。用一个变量记住下次出现的元素需要大于还是需要小于前一个元素。也可以用模拟的方法，用两个变量分别记录上升和下降数字的长度。一旦元素相等了，上升和下降数字长度都置为 1，其他时候按照上升和下降的关系增加队列长度即可，最后输出动态维护的最长长度。