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.

使用二分搜索需计算l,r 左右边界

1.当A[l] <= A[m]时 说明左半部分已经排好序

2.当A[m] < A[r]时 说明右半部分已经排好序

public int search(int[] A, int target) {
        int len = A.length;
        
        // binary search
        int l = 0;
        int r = len - 1;
        while(l <= r){
            int m = (l + r) / 2;
            if(target == A[m])
                return m;
            
            // lower is sorted
            if(A[l] <= A[m]){
                if(A[l] <= target && target < A[m])
                    r = m - 1;
                else{
                    l = m + 1;
                }
            } else {
                if(A[m] < target && target <= A[r]){
                    l = m + 1;
                } else{
                    r = m - 1;
                }
            }
        }
        return -1;
    }