feat: 33. Search in Rotated Sorted Array

Author: kimi0230

Leetcode/0033.Search-in-Rotated-Sorted-Array/0033.Search-in-Rotated-Sorted-Array.jpg

@@ -0,0 +1,174 @@
+---
+title: 33. Search in Rotated Sorted Array
+subtitle: "https://leetcode.com/problems/search-in-rotated-sorted-array/description/"
+date: 2024-03-31T16:26:00+08:00
+lastmod: 2024-03-31T16:26:00+08:00
+draft: false
+author: "Kimi.Tsai"
+authorLink: "https://kimi0230.github.io/"
+description: "0033.Search-in-Rotated-Sorted-Array"
+license: ""
+images: []
+
+tags: [LeetCode, Go, Medium, Array, Binary Search, 33. Search in Rotated Sorted Array, Amazon,Facebook, Microsoft, LinkedIn, Apple]
+categories: [LeetCode]
+
+featuredImage: ""
+featuredImagePreview: ""
+
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+hiddenFromSearch: false
+twemoji: false
+lightgallery: true
+ruby: true
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+linkToMarkdown: false
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+
+toc:
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+  auto: true
+code:
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+  maxShownLines: 200
+math:
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+    # Or
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+  js:
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+    # located in "assets/"
+    # Or
+    # someJS = "https://cdn.example.com/some.js"
+seo:
+  images: []
+  # ...
+---
+# [33. Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/description/)
+
+## 題目
+There is an integer array nums sorted in ascending order (with distinct values).
+
+Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
+
+Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
+
+You must write an algorithm with O(log n) runtime complexity.
+
+ 
+
+Example 1:
+
+Input: nums = [4,5,6,7,0,1,2], target = 0
+Output: 4
+Example 2:
+
+Input: nums = [4,5,6,7,0,1,2], target = 3
+Output: -1
+Example 3:
+
+Input: nums = [1], target = 0
+Output: -1
+ 
+
+Constraints:
+
+1 <= nums.length <= 5000
+-104 <= nums[i] <= 104
+All values of nums are unique.
+nums is an ascending array that is possibly rotated.
+-104 <= target <= 104
+
+## 題目大意
+整個數組按照升序排序, 值不相同
+在傳遞給函數之前， nums 在預先未知的某個下標 k （ 0 <= k < nums.length ）上進行了 旋轉，使數位變數 [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] （下標 從 0 開始 計數）。 例如， [0,1,2,4,5,6,7] 在下標 3 處經旋轉後可能變為 [4,5,6,7,0,1,2] 。
+
+給你 旋轉後 的陣列 nums 和一個整數 target ，如果 nums 中存在這個目標值 target ，則返回它的下標，否則返回 -1 。
+
+你必須設計一個時間複雜度為 O(log n) 的演演算法解決此問題。
+
+## 解題思路
+
+先判斷mid位於上半部還是下半部, 再來找target位於哪邊
+最後看target是否有在左或右端點
+
+![](0033.Search-in-Rotated-Sorted-Array.jpg)
+
+
+## Big O
+
+* 時間複雜 : `O(log n)`
+* 空間複雜 : `O(1)`
+
+## 來源
+* https://leetcode.com/problems/search-in-rotated-sorted-array/description/
+* https://leetcode.cn/problems/
+
+## 解答
+https://github.com/kimi0230/LeetcodeGolang/blob/master/Leetcode/0033.Search-in-Rotated-Sorted-Array/main.go
+
+```go
+package searchinrotatedsortedarray
+
+// 時間複雜 O(), 空間複雜 O()
+func search(nums []int, target int) int {
+	if nums == nil || len(nums) == 0 {
+		return -1
+	}
+	left, right := 0, len(nums)-1
+	for left < right {
+		mid := int(uint(left+right) >> 1)
+
+		if nums[mid] == target {
+			return mid
+		}
+
+		if nums[mid] >= nums[left] {
+			// mid在上半區
+
+			if nums[left] <= target && target <= nums[mid] {
+				// target 在left跟min中間
+				right = mid - 1
+			} else {
+				left = mid + 1
+			}
+		} else {
+			// mid在下半區
+			if nums[mid] <= target && target <= nums[right] {
+				// target 在min跟right中間
+				left = mid + 1
+			} else {
+				right = mid - 1
+			}
+		}
+	}
+	if nums[left] == target {
+		return left
+	} else if nums[right] == target {
+		return right
+	} else {
+		return -1
+	}
+}
+
+```
+
+##  Benchmark
+
+```sh
+
+```

Leetcode/0033.Search-in-Rotated-Sorted-Array/README.md

@@ -0,0 +1,42 @@
+package searchinrotatedsortedarray
+
+// 時間複雜 O(), 空間複雜 O()
+func search(nums []int, target int) int {
+	if nums == nil || len(nums) == 0 {
+		return -1
+	}
+	left, right := 0, len(nums)-1
+	for left < right {
+		mid := int(uint(left+right) >> 1)
+
+		if nums[mid] == target {
+			return mid
+		}
+
+		if nums[mid] >= nums[left] {
+			// mid在上半區
+
+			if nums[left] <= target && target <= nums[mid] {
+				// target 在left跟min中間
+				right = mid - 1
+			} else {
+				left = mid + 1
+			}
+		} else {
+			// mid在下半區
+			if nums[mid] <= target && target <= nums[right] {
+				// target 在min跟right中間
+				left = mid + 1
+			} else {
+				right = mid - 1
+			}
+		}
+	}
+	if nums[left] == target {
+		return left
+	} else if nums[right] == target {
+		return right
+	} else {
+		return -1
+	}
+}

Leetcode/0033.Search-in-Rotated-Sorted-Array/main.go

@@ -0,0 +1,36 @@
+package searchinrotatedsortedarray
+
+import (
+	"testing"
+)
+
+var tests = []struct {
+	nums   []int
+	target int
+	want   int
+}{
+	{[]int{4, 5, 6, 7, 0, 1, 2}, 0, 4},
+	{[]int{4, 5, 6, 7, 0, 1, 2}, 3, -1},
+	{[]int{1}, 0, -1},
+}
+
+func TestSearch(t *testing.T) {
+	for _, tt := range tests {
+		if got := search(tt.nums, tt.target); got != tt.want {
+			t.Errorf("search(%v, %v) = %v, want %v", tt.nums, tt.target, got, tt.want)
+		}
+	}
+}
+
+func BenchmarkSearch(b *testing.B) {
+	for i := 0; i < b.N; i++ {
+		for _, tt := range tests {
+			search(tt.nums, tt.target)
+		}
+	}
+}
+
+/*
+go test -benchmem -run=none LeetcodeGolang/Leetcode/0354.Russian-Doll-Envelopes -bench=.
+
+*/

Leetcode/0033.Search-in-Rotated-Sorted-Array/main_test.go

@@ -0,0 +1,51 @@
+// 時間複雜度:
+// 空間複雜度:
+/*
+ * @lc app=leetcode.cn id=33 lang=golang
+ *
+ * [33] 搜索旋转排序数组
+ */
+
+// @lc code=start
+func search(nums []int, target int) int {
+	if nums == nil || len(nums) == 0 {
+		return -1
+	}
+	left, right := 0, len(nums)-1
+	for left < right {
+		mid := int(uint(left+right) >> 1)
+
+		if nums[mid] == target {
+			return mid
+		}
+
+		if nums[mid] >= nums[left] {
+			// mid在上半區
+
+			if nums[left] <= target && target <= nums[mid] {
+				// target 在left跟min中間
+				right = mid - 1
+			} else {
+				left = mid + 1
+			}
+		} else {
+			// mid在下半區
+			if nums[mid] <= target && target <= nums[right] {
+				// target 在min跟right中間
+				left = mid + 1
+			} else {
+				right = mid - 1
+			}
+		}
+	}
+	if nums[left] == target {
+		return left
+	} else if nums[right] == target {
+		return right
+	} else {
+		return -1
+	}
+}
+
+// @lc code=end
+

Leetcode_labuladong/33.搜索旋转排序数组.go

@@ -487,6 +487,7 @@ func RightBound(nums []int, target int) (index int) {
 | [0704](https://kimi0230.github.io/LeetcodeGolang/Leetcode/0704.Binary-Search/)                        |                              [704. Binary Search](https://leetcode.com/problems/binary-search/)                              |            [Go](https://github.com/kimi0230/LeetcodeGolang/tree/master/Leetcode/0704.Binary-Search)             | Easy       | 最差:O(long n)<br> 最佳O(1)剛好在中間 | 迭代: O(1) <br/> 遞迴O(log n) | Binary Search |
 | [0875](https://kimi0230.github.io/LeetcodeGolang/Leetcode/0875.Koko-Eating-Bananas/)                  |                  [875. Koko Eating Bananas](https://leetcode.com/problems/koko-eating-bananas/description/)                  |         [Go](https://github.com/kimi0230/LeetcodeGolang/tree/master/Leetcode/0875.Koko-Eating-Bananas)          | Medium     | O（n log m）                          | O(1)                          | Binary Search |
 | [0153](https://kimi0230.github.io/LeetcodeGolang/Leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/) | [153. Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/) | [Go](https://github.com/kimi0230/LeetcodeGolang/tree/master/Leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array) | Medium     | O(log n）                             | O(1)                          | Binary Search |
+| [0033](https://kimi0230.github.io/LeetcodeGolang/Leetcode/0033.Search-in-Rotated-Sorted-Array/)       |       [33. Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/description//)       |    [Go](https://github.com/kimi0230/LeetcodeGolang/tree/master/Leetcode/0033.Search-in-Rotated-Sorted-Array)    | Medium     | O(log n）                             | O(1)                          | Binary Search |
 
 
 ---