81. Search in Rotated Sorted Array II

题目：

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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

链接： http://leetcode.com/problems/search-in-rotated-sorted-array-ii/

题解：

平移过的数组查找数字。依然是使用二分查找，不过有了duplicate所以判断的条件更多更复杂。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {public boolean search(int[] nums, int target) {if(nums == null || nums.length == 0)return false;int lo = 0, hi = nums.length - 1;while(lo <= hi) {int mid = lo + (hi - lo) / 2;if(target < nums[mid]) {if(nums[mid] < nums[hi]) {      // right half sortedhi = mid - 1;} else if(nums[mid] > nums[hi]) {   //left half sorted if(target < nums[lo]) {lo = mid + 1;           //target at right half} else {hi = mid - 1;           //target at left half
                    }} else {                        // nums[mid] == nums[hi], cannot tellhi--;}} else if (target > nums[mid]) {if(nums[lo] < nums[mid]) {       //left half sortedlo = mid + 1;} else if(nums[lo] > nums[mid]) { // right half sortedif(target > nums[hi])hi = mid - 1;           //target at left halfelselo = mid + 1;           //target at right half} else {                          // nums[mid] == nums[hi], cannot telllo++;}} else                      // nums[mid] == target, foundreturn true;}return false;}
}

二刷：

还是使用二分法来完成搜索。这里不同的地方是，假如nums[mid]等于一个端点的时候，假如是左端点，那么我们不能判断出哪一边是排序的，只能lo++。否则，我们可以通过比较判断出左边或者右边是排序的，从而使用二分查找。

Java:

Time Complextiy - O(n)， Space Complexity - O(1)

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

2/6/2016

题外话：

国内今天已经是29号，除夕了。现在我一个人在美帝的家里，中午刚上完课，下午也参加了一个tech talk。依然是平实的一天，就是觉得有点蛋疼，该春节回家陪陪爸妈的。我打算明天给自己放假一天，好好看会电视，然后去H-Mart买桶炸鸡吃掉。

三刷:

其实29号那天并没有吃炸鸡。

注意比较的都是nums[lo]和nums[mid]。我们在比较晚nums[lo]和nums[mid]之后，再假定一边有序，一边无序的情况。假如nums[lo] == nums[lo]，我们不能二分，只能增加lo来继续下一次判断。

Java:

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

Reference:

https://leetcode.com/discuss/223/when-there-are-duplicates-the-worst-case-is-could-we-do-better

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