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++

public class Solution {
    public boolean search(int[] A, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len = A.length;
        
        // binary search
        int l = 0;
        int r = len - 1;
        while(l <= r){
            int m = (l + r) / 2;
            if(target == A[m])
                return true;
            
            // lower is sorted
            if(A[l] < A[m]){
                if(A[l] <= target && target < A[m])
                    r = m - 1;
                else{
                    l = m + 1;
                }
            } else if(A[l] > A[m]) {
                // upper is sorted
                if(A[m] < target && target <= A[r]){
                    l = m + 1;
                } else{
                    r = m - 1;
                }
            } else {
                l = l + 1;
            }
        }
        return false;
    }
}

 

ref:

二分搜索及其扩展

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