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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 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.

思路分析:这题同样考察二分查找,但是是在rotation之后的数组中进行二分查找。这题的特点是,每次二分后,至少有一半是有序的,另一半是无序的(受到了rotation的影响)。所以要利用这一特性,通过比较target和有序的那半边数组的最小数和最大数,来决定l和m的更新规则。具体而言

1.如果target = A[m] 返回m

2 如果target 不等于A[m]

2.1 如果A[m] < A[r],说明A[m...r]有序,没有收到rotation影响,可以判断target是否在A[m...r]范围内,如果是更新l=m+1;否则更新r=m-1

2.2 如果A[m] >= A[r],说明A[l...m]有序,没有收到rotation影响,可以判断target是否在A[l...m]范围内,如果是更新r=m-1;否则更新l=m+1

所以利用每次仍然有序的半边数组,我们仍然可以进行二分查找,更新l和r,每次缩小一半的搜索范围,时间复杂度O(log(n)),空间复杂度O(1),只有常数级别的额外空间。注意判断target是否在A[m...r]范围和target是否在A[l...m]范围内的时候,要考虑target == A[r]和target == A[l]的情况,二分查找的实现等号情况的考虑要特别小心。

AC 时间:5min。

AC Code

public class Solution {
    public int search(int[] A, int target) {
        //02:46
        if(A == null || A.length == 0) return -1;
        
        int l = 0;
        int r = A.length - 1;
        
        while(l <= r){
            int m = (l + r) / 2;
            if(target == A[m]) return m;
            else{
                if(A[m] < A[r]){
                    //m ... r is sorted
                    if(target <= A[r] && target > A[m]) {
                        l = m + 1;
                    } else {
                        r = m - 1;
                    }
                } else {
                    //l ... m is sorted
                    if(target >= A[l] && target < A[m]){
                        r = m -1;
                    } else {
                        l = m + 1;
                    }
                }
            }
        }
        return -1;
    }
    //02:51
}


LeetCode Search in Rotated Sorted Array