leetcode -- Search in Rotated Sorted Array II

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.

上题中A[l] <= A[m]在数组中有重复元素时无法保证[l,m]是sorted

当输入：[1,3,1,1,1], 3

这里参考：

分为两种情形：http://fisherlei.blogspot.com/2013/01/leetcode-search-in-rotated-sorted-array_3.html

1. A[l] < A[m] 则[l, m]是递增的

2. A[l] = A[m] 此时是重复元素，无法确认哪一部分是有序的，则跳过l，l++

 1 public class Solution {
 2     public boolean search(int[] A, int target) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int len = A.length;
 6         
 7         // binary search
 8         int l = 0;
 9         int r = len - 1;
10         while(l <= r){
11             int m = (l + r) / 2;
12             if(target == A[m])
13                 return true;
14             
15             // lower is sorted
16             if(A[l] < A[m]){
17                 if(A[l] <= target && target < A[m])
18                     r = m - 1;
19                 else{
20                     l = m + 1;
21                 }
22             } else if(A[l] > A[m]) {
23                 // upper is sorted
24                 if(A[m] < target && target <= A[r]){
25                     l = m + 1;
26                 } else{
27                     r = m - 1;
28                 }
29             } else {
30                 l = l + 1;
31             }
32         }
33         return false;
34     }
35 }

 

ref:

二分搜索及其扩展

http://www.pcw8510.com/?p=3294