Search in Rotated Sorted Array II LeetCode Java

描述
Follow up for ”Search in Rotated Sorted Array”: What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
分析
允许重复元素，则上一题中如果 A[m]>=A[l], 那么 [l,m] 为递增序列的假设就不能成立了，比
如 [1,3,1,1,1]。
如果 A[m]>=A[l] 不能确定递增，那就把它拆分成两个条件：
• 若 A[m]>A[l]，则区间 [l,m] 一定递增
• 若 A[m]==A[l] 确定不了，那就 l++，往下看一步即可。

代码

 1 // LeetCode, Search in Rotated Sorted Array II
 2 // 时间复杂度 O(log n)，空间复杂度 O(1)
 3 class Solution {
 4              public static boolean 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 true;
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 if (A[first] > A[mid]) {
16                                   if (A[mid] <= target && target <= A[last-1])
17                                              first = mid + 1;
18                                    else
19                                              last = mid;
20                                   } else
21                              //skip duplicate one, A[start] == A[mid]
22                                               first++;
23                      }
24                       return false;
25             }
26 }