元素出现区间(两次二分搜索)

Author: cls1991

BinarySearch/34_SearchForARange.py

@@ -0,0 +1,66 @@
+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/search-for-a-range/description.
+    题目描述:
+
+    Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.
+
+    Your algorithm's runtime complexity must be in the order of O(log n).
+
+    If the target is not found in the array, return [-1, -1].
+
+    For example,
+    Given [5, 7, 7, 8, 8, 10] and target value 8,
+    return [3, 4].
+
+"""
+
+
+class Solution(object):
+    def searchRange(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+        if not nums:
+            return [-1, -1]
+
+        # 两次二分搜索
+        return [self.searchLeftRange(nums, target), self.searchRightRange(nums, target)]
+
+    def searchLeftRange(self, nums, target):
+        left = 0
+        right = len(nums) - 1
+        while left <= right:
+            mid = left + (right - left) // 2
+            if nums[mid] == target:
+                right = mid - 1
+            elif nums[mid] < target:
+                left = mid + 1
+            else:
+                right = mid - 1
+
+        if left < len(nums) and nums[left] == target:
+            return left
+
+        return -1
+
+    def searchRightRange(self, nums, target):
+        left = 0
+        right = len(nums) - 1
+        while left <= right:
+            mid = left + (right - left) // 2
+            if nums[mid] == target:
+                left = mid + 1
+            elif nums[mid] < target:
+                left = mid + 1
+            else:
+                right = mid - 1
+
+            if right >= 0 and nums[right] == target:
+                return right
+
+        return -1