leetcode_81. Search in Rotated Sorted Array II

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

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

给一个有重复值、升序排列但被旋转过的数组，找出有没有指定值

第一种，麻烦且不高级。。

class Solution {
    public boolean search(int[] nums, int target) {
        if (nums.length < 1)
            return false;
        int start = 0, end = nums.length-1;
        if (target > nums[start]){
            for (int i = 0; i < nums.length; i++){
                if (nums[i] == target)
                    return true;
            }
        }
        else if (target < nums[end]){
            for (int i = end; i >= 0; i--){
                if (nums[i] == target)
                    return true;
            }
        }
        return target==nums[start] || target==nums[end];
    }
}

二分查找：同样是讨论区看到的答案，类似二分查找，在代码中加了一些注释，有什么问题欢迎留言交流

class Solution {
    public boolean search(int[] nums, int target) {
       int low = 0, hi = nums.length - 1, mid =-1;
        while (low <= hi){
            mid = low + (hi - low)/2;
            if (nums[mid] == target)
                return true;
            // 当数组左边是升序序列
            if (nums[low] < nums[mid]){
                // 判断target位置
                if (target > nums[mid] || target < nums[low])
                    low = mid+1;
                else hi = mid-1;
            }// 如果数组右边是升序序列
             else if (nums[low] > nums[mid]){
                if (target < nums[mid] || target > nums[hi])
                    hi = mid-1;
                else low = mid+1;
            }else {
                low++;
            }
        }
        return false;
    }
}