lintcode-medium-Search in Rotated Sorted Array II-优快云博客

本文提供了一种在允许重复元素的旋转排序数组中查找特定目标值的方法。通过递归二分搜索策略，即使数组中有重复元素也能有效地找到目标值。

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.

public class Solution {
    /** 
     * param A : an integer ratated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean 
     */
    public boolean search(int[] A, int target) {
        // write your code here
        
        if(A == null || A.length == 0)
            return false;
        
        int left = 0;
        int right = A.length - 1;
        
        return search(A, target, left, right);
    }
    
    public boolean search(int[] A, int target, int left, int right){
        
        if(left > right)
            return false;
        
        if(left == right){
            if(A[left] == target)
                return true;
            else
                return false;
        }
        
        int mid = left + (right - left) / 2;
        
        if(A[mid] == target)
            return true;
            
        if(A[mid] > target){
            if(A[mid] > A[left]){
                boolean flag1 = search(A, target, left, mid - 1);
                boolean flag2 = search(A, target, mid + 1, right);
                    
                if(!flag1 && !flag2)
                    return false;
                else if(!flag1)
                    return flag2;
                else
                    return flag1;
            }
            else if(A[mid] < A[left]){
                return search(A, target, left, mid - 1);
            }
            else{
                return search(A, target, left + 1, right);
            }
        }
        else{
            if(A[mid] > A[left]){
                return search(A, target, mid + 1, right);
            }
            else if(A[mid] < A[left]){
                boolean flag1 = search(A, target, left, mid - 1);
                boolean flag2 = search(A, target, mid + 1, right);
                    
                if(!flag1 && !flag2)
                    return false;
                else if(!flag1)
                    return flag2;
                else
                    return flag1;
            }
            else{
                return search(A, target, left + 1, right);
            }
        }
        
    }

}