33. Search in Rotated Sorted Array

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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.

解法1：

Binary Search的经典题。主要是要考虑怎么扔掉一半。
C++

1

Java

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