旋转数组中二分搜索

33. Search in Rotated Sorted Array/main.py

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+"""
+非常复杂，思考的方向错了。应该首先判断mid
+"""
+
+
+class Solution:
+    def search(self, nums, target) -> int:
+        left, right = 0, len(nums) - 1
+        mid = 0
+
+        while left <= right:
+            mid = (left + right) // 2
+            # if nums[left]<target<nums[mid]:
+            #     right = mid
+            if target > nums[left]:
+                if nums[mid] > target:
+                    # nums[left] < target < nums[mid]
+                    right = mid - 1
+                elif nums[mid] < nums[left]:
+                    # nums[mid] < nums[left] < target
+                    right = mid - 1
+                elif nums[mid] < target:
+                    # nums[left] <= nums[mid]<=target
+                    left = mid + 1
+                else:
+                    return mid
+
+            elif target < nums[left]:
+                if nums[mid] < target:
+                    # nums[mid] < target < nums[left]
+                    left = mid + 1
+                elif nums[mid] > nums[left]:
+                    # target<nums[left]<nums[mid]
+                    left = mid + 1
+                elif nums[mid] > target:
+                    # target<nums[mid]<=nums[left]
+                    right = mid - 1
+                else:
+                    return mid
+            else:
+                return left
+
+        return -1
+
+
+nums = [4, 5, 6, 7, 0, 1, 2]
+target = 0
+nums = [1]
+target = 0
+print(Solution().search(nums, target))

33. Search in Rotated Sorted Array/normal.go

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+package problem33
+
+func search(nums []int, target int) int {
+	if len(nums) == 0 {
+		return -1
+	}
+
+	left, right := 0, len(nums)-1
+	var mid int
+	for left < right {
+		mid = (left + right) / 2
+		if nums[mid] < nums[left] {
+			if target > nums[mid] && target <= nums[right] {
+				// 为什么left要mid+1，而right可以=mid？
+				// 因为mid=(left+right)/2,mid会=left，所以要+1
+				left = mid + 1
+			} else {
+				right = mid
+			}
+		} else { // nums[mid] >= nums[left]
+			if nums[left] <= target && target <= nums[mid] {
+				right = mid
+			} else {
+				left = mid + 1
+			}
+		}
+	}
+
+	if nums[left] == target {
+		return left
+	}
+	return -1
+}

33. Search in Rotated Sorted Array/others.org

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+* other solutions
+
+** find minimum
+Firstly, find the minimum. Then, see the List as a ascending list which binary search can be applied to.
+Time complexity: O(logN).
+#+BEGIN_SRC python
+
+class Solution:
+    def search(self, nums: 'List[int]', target: 'int') -> 'int':
+        left = 0
+        right = len(nums) - 1
+        while left < right:
+            mid = int((left + right)/2)
+            if nums[mid] < nums[right]:
+                right = mid
+            elif nums[mid] > nums[right]:
+                left = mid + 1
+        print(left)
+        lo = left
+        left = 0
+        right = len(nums) - 1
+        while left <= right:
+            mid = int((left + right)/2)
+            real_mid = (mid + lo ) % len(nums)
+            if nums[real_mid] == target:
+                return real_mid
+            if nums[real_mid] < target:
+                left = mid + 1
+            elif nums[real_mid] > target:
+                right = mid - 1
+
+        return -1
+
+#+END_SRC
+
+** if
+
+py soln 看到的。用 go 重写。
+#+BEGIN_SRC golang
+
+func search(nums []int, target int) int {
+	left, right := 0, len(nums)-1
+	var mid int
+	for left <= right {
+		mid = (left + right) / 2
+		if nums[mid] == target {
+			return mid
+		}
+		switch {
+		case target < nums[0] && nums[0] <= nums[mid],
+			nums[0] <= nums[mid] && nums[mid] < target,
+			nums[mid] < target && target <= nums[len(nums)-1]:
+			left = mid + 1
+		default:
+			right = mid - 1
+		}
+	}
+	return -1
+}
+
+#+END_SRC

33. Search in Rotated Sorted Array/readme.org

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+#+TITLE: 文件说明。
+* O(N)
+The solution with O(N) time complexity and that with O(logN) cost almost the same time.
+
+It seems that time used in leetcode does not have reference value.
+
+* complex
+main.py
+情况太多。复杂到自己都搞不懂，放弃分析。
+也可能是因为自己笨，没有化繁为简的能力。
+
+* 标准解法
+normal.go
+
+* others
+其他解法