33. Search in Rotated Sorted Array I & II

Suppose a sorted array 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.

 

Example

For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

分析：

二分搜索。需要把升序部分和target进行比较。

public class Solution {

    public int search(int[] A, int target) {
        if (A == null || A.length == 0) return -1;
        int left = 0, right = A.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (A[mid] == target) {
                return mid;
            } else if (A[left] <= A[mid]) { // we have to use <= to handle the case in which we have [3, 1] and target is 1.
                if (A[left] <= target && target < A[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else {
                if (A[mid] < target && target <= A[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }
        return -1;
    }
}

 

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.

 

Example

Given [1, 1, 0, 1, 1, 1] and target = 0, return true.
Given [1, 1, 1, 1, 1, 1] and target = 0, return false.

分析：

用一个方法来确定其中的一半是否单调递增。

class Solution {
    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(nums[start] <= target && target < nums[mid]){
                    end = mid - 1;
                }else{
                    //target is in rotated part
                    start = mid + 1;
                }
            }else if(nums[start] > nums[mid]){
                //right part is sorted
                //target is in rotated part
                if(target > nums[mid] && target <= nums[end]){
                    start = mid + 1;
                }else{
                    end = 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;
    }
}