Search in Rotated Sorted Array ||

Search in Rotated Sorted Array ||

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.

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.

ref:http://fisherlei.blogspot.com/2013/01/leetcode-search-in-rotated-sorted-array_3.html

[解题思路]
确实有影响。比如，上一题(http://fisherlei.blogspot.com/2013/01/leetcode-search-in-rotated-sorted-array.html)的程序中默认如果A[m]>=A[l],那么[l,m]为递增序列的假设就不能成立了，比如如下数据
[1,3,1,1,1]
所以，要是想增强该假设，有两个选择
1. 对于每一个递增序列，遍历之，确认。
2. 找到pivot点，然后确定对应序列搜索。


Update: 3/18/2013. add the implementation
重新想了一下，其实不需要这么复杂。如果A[m]>=A[l]不能确定递增，那就把它拆分成两个条件
1. A[m]>A[l]  递增
2. A[m] ==A[l] 确定不了，那就l++，往下看一步即可。 

public class Solution {
    public boolean search(int[] A, int target) {
        int len = A.length;
        
        int l = 0;
        int r = len -1;
        
        while(l <= r){
            int m = (l+r)/2;
            
            if(target == A[m]){
                return true;
            }
            
            if( A[m] < A[l]){ // higher is sorted
                if(target > A[m] && target <= A[r]){
                    l = m+1;
                }else{
                    r = m-1;
                }
            }else if(A[m] > A[l]){
                if(target >= A[l] && target < A[m]){
                    r = m-1;
                }else{
                    l = m+1;
                }
            }else{  // A[m] == A[l]
                    l++;
            }
        }
        
        return false;
    }
}