81. Search in Rotated Sorted Array II

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 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

题目大意：给定一个旋转后的数组，写一个函数，判断给定的数，是否在数组中。

思路：

直接遍历数组，判断是否有target，时间复杂度O（N）。

二分查找，查找目标数，时间复杂度O（logN）.

代码：

package Array;

/**
* @Author OovEver
* @Date 2017/11/28 10:10
*/
public class Solution {
public boolean search(int[] nums, int target) {
int start = 0, end = nums.length - 1, mid = -1;
while(start <= end) {
mid = (start + end) / 2;
if (nums[mid] == target) {
return true;
}
//If we know for sure right side is sorted or left side is unsorted
if (nums[mid] < nums[end] || nums[mid] < nums[start]) {
if (target > nums[mid] && target <= nums[end]) {
start = mid + 1;
} else {
end = mid - 1;
}
//If we know for sure left side is sorted or right side is unsorted
} else if (nums[mid] > nums[start] || nums[mid] > nums[end]) {
if (target < nums[mid] && target >= nums[start]) {
end = mid - 1;
} else {
start = mid + 1;
}
//If we get here, that means nums[start] == nums[mid] == nums[end], then shifti
4000
ng out
//any of the two sides won't change the result but can help remove duplicate from
//consideration, here we just use end-- but left++ works too
} else {
//                low++也可以
end--;
}
}

return false;
}

}

​