Search in Rotated Sorted Array

题目
Suppose an array sorted in ascending order 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.

答案

方法1

二分思路

If nums[mid] >= nums[start], we know that nums[start…mid] is definitely sorted
    If target is inside start…mid, that would be great, we can just perform a binary search on the interval start…mid

    If target is outside of start…mid, meaning it could be greater than nums[mid] or it could be smaller than nums[start]
    In this case we reduce the search space from start…end to mid + 1…end

If nums[mid] < nums[start], we know that nums[start…mid] is definitely not sorted, which means nums[mid+1…end] is definitely sorted
    If target is inside mid+1…end, that would be great, we can just perform a binary search on the interval mid+1…end

    If target is outside of mid+1…end, meaning it could be greater than nums[end] or it could be smaller than nums[mid+1]
    In this case we reduce the search space from start…end to start…mid - 1

代码

class Solution {
    public int search(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while(l <= r) {
            int m = (l + r) / 2;
            if(target == nums[m]) return m;

            if(nums[l] <= nums[m]) {
                if(target >= nums[l] && target <= nums[m]) {
                    r = m - 1;
                }
                else {
                    l = m + 1;
                }
            }
            else {
                if(target >= nums[m] && target <= nums[r]) {
                    l = m + 1;
                }
                else {
                    r = m - 1;
                }
            }
        }
        return -1;
    }
}

方法2
先用binary search找最小元素的index，以这个index把array分成两边，然后对两边分别做binary search