LeetCode(Sorting&Searching)Search a 2D Matrix II

Search in Rotated Sorted Array

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.

Your algorithm’s runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

思路：最简单的办法，穷举，时间复杂度o(n)。

代码：

class Solution {
    public int search(int[] nums, int target) {
        for(int i=0; i<nums.length; i++){
            if(nums[i]==target){
                return i;
            }
        }
        return -1;
    }
}