81. Search in Rotated Sorted Array II

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

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

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.

在rotated的数组里查找目标值，数组包含重复元素，这时分三种情况来判断，详见代码：

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 shifting 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 {            end--;        }    }        return false;}这种方法需要检验两端，下面的方法不需要检验两端：
public boolean search(int[] nums, int target) {    int start  = 0, end = nums.length - 1;        //check each num so we will check start == end    //We always get a sorted part and a half part    //we can check sorted part to decide where to go next    while(start <= end){        int mid = start + (end - start)/2;        if(nums[mid] == target) return true;                //if left part is sorted        if(nums[start] < nums[mid]){            if(target < nums[start] || target > nums[mid]){                //target is in rotated part                start = mid + 1;            }else{                end = mid - 1;            }        }else if(nums[start] > nums[mid]){            //right part is sorted                        //target is in rotated part            if(target < nums[mid] || target > nums[end]){                end = mid -1;            }else{                start = mid + 1;            }        }else{            //duplicates, we know nums[mid] != target, so nums[start] != target            //based on current information, we can only move left pointer to skip one cell            //thus in the worest case, we would have target: 2, and array like 11111111, then            //the running time would be O(n)            start ++;        }    }        return false;}