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。

二分查找

代码

// LeetCode, Search in Rotated Sorted Array II
// 时间复杂度 O(log n)，空间复杂度 O(1)
class Solution {
             public static boolean search(int A[], int n, int target) {
                     int first = 0, last = n;
                     while (first != last) {
                             int mid = (first + last) / 2;
                             if (A[mid] == target)
                             return true;
                             if (A[first] < A[mid]) {
                                   if (A[first] <= target && target < A[mid])
                                             last = mid;
                                    else
                                             first = mid + 1;
                            } else if (A[first] > A[mid]) {
                                  if (A[mid] <= target && target <= A[last-1])
                                             first = mid + 1;
                                   else
                                             last = mid;
                                  } else
                             //skip duplicate one, A[start] == A[mid]
                                              first++;
                     }
                      return false;
            }
}