标签：int duplicate 难度 sea return stat log bec solution
描述
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
分析
二分查找，难度主要在于左右边界的确定。
代码
1 // LeetCode, Search in Rotated Sorted Array 2 // 时间复杂度 O(log n)，空间复杂度 O(1) 3 class Solution { 4 public static int search(int A[], int n, int target) { 5 int first = 0, last = n; 6 while (first != last) { 7 int mid = (first + last) / 2; 8 if (A[mid] == target) 9 return mid; 10 if (A[first] <= A[mid]) { 11 if (A[first] <= target && target < A[mid]) 12 last = mid; 13 else 14 first = mid + 1; 15 } else { 16 if (A[mid] < target && target <= A[last-1]) 17 first = mid + 1; 18 else 19 last = mid; 20 } 21 } 22 return -1; 23 } 24 }
Search in Rotated Sorted Array leetcode java
标签：int duplicate 难度 sea return stat log bec solution
原文地址：https://www.cnblogs.com/ncznx/p/9167398.html
