[leetcode] 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 7might become4 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.

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

思路：使用二分查找，若中间数大于等于最左端数，那左半部分必定有序，否则右半部分有序，通过将目标数与有序部分的最大与最小数比较来判断目标数位于哪一部分。复杂度O(logn)。

注意：此题边界情况，>,>=等需仔细考虑斟酌。

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

public class Solution {
    public int search(int[] A, int target) {
        if (A == null || A.length == 0)
            return -1;

        int l = 0, r = A.length - 1, mid;
        while (l <= r) {
            mid = (l + r) / 2;
            if (A[mid] == target)
                return mid;
            else if (A[l] <= A[mid]) {
                if (A[l] <= target && target < A[mid])
                    r = mid - 1;
                else
                    l = mid + 1;

            } else {
                if (A[mid] < target && target <= A[r])
                    l = mid + 1;
                else
                    r = mid - 1;

            }

        }
        return -1;

    }

    public static void main(String[] args) {
        System.out.println(new Solution().search(new int[] { 3,1 }, 1));

        System.out.println(new Solution().search(new int[] { 1, 2, 3, 4, 5 }, 3));
        System.out.println(new Solution().search(new int[] { 2, 3, 4, 1, 2 }, 3));
        System.out.println(new Solution().search(new int[] { 2, 3, 4, 1, 2 }, 1));
        System.out.println(new Solution().search(new int[] { 4, 5, 1, 2, 3 }, 5));
        System.out.println(new Solution().search(new int[] { 4, 5, 1, 2, 3 }, 3));

    }

}

第二遍记录：

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

}

参考：

http://blog.csdn.net/zjull/article/details/11780329

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