[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.

https://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/

 

思路：类似Search in Rotated Sorted Array的思路，只不过现在的问题是有重复，可能出现a[l]==a[m]的极端情况，无法判断哪半边有序，因此这种情况下只能一步一步移动l 直到不相等然后通过二分判断。

复杂度：极端情况每次都是减一 而不是减去一半，所以复杂度是O(N)。

 

/**
 * http://blog.csdn.net/linhuanmars/article/details/20525681
 * 
 * @author jd
 * 
 */

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

    public static void main(String[] args) {

        System.out.println(new Solution().search(new int[] { 3, 1, 1 }, 3));

        System.out.println(new Solution().search(new int[] { 2, 3, 3, 3, 3 }, 2));
        System.out.println(new Solution().search(new int[] { 3, 2, 3, 3, 3 }, 2));
        System.out.println(new Solution().search(new int[] { 3, 3, 3, 2, 3 }, 2));
        System.out.println(new Solution().search(new int[] { 3, 3, 3, 3, 2 }, 2));

    }

}

 

第二遍记录：

相比没有重复元素的，当A[left]==A[mid]的时候，只能一步一步移动left直到不相等。

public class Solution {
    public boolean search(int[] A, int target) {
        if(A==null||A.length==0)
            return false;
        int left=0,right=A.length-1,mid=0;
        while(left<=right){
            mid = left+(right-left)/2;
            if(target==A[mid])
                return true;
            if(A[left]<A[mid]){
                if(target>=A[left]&&target<A[mid]){
                    right=mid-1;
                }else{
                    left=mid+1;
                }
            }else if(A[left]>A[mid]){
                if(target>A[mid]&&target<=A[right]){
                    left=mid+1;
                }else{
                    right=mid-1;
                }
                
            }else{
                left++;
            }
            
        }
        return false;
    }

}

 

 

参考：

http://blog.csdn.net/linhuanmars/article/details/20525681