33. Search in Rotated Sorted Array

Problem:

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]).

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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Analysis:
6/19/2018 update:
If mid value is on upper part, it's easier to control going left. So that target has to on the upper part: target >= nums[left]. and target needs to be smaller than mid value, target < nums[mid]. Same thing applies to lower part.

The left and right points are inclusive. So add == to condition.

No need to use start + 1 < end template, since this problem finds the exact value.

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

_________________________________________________________________________________

01/16/2018 update

if (nums[start] < target < nums[mid]) move to left
if (nums[mid] < target < nums[end]) move to right
--------------------------------------------------------------------------------------------------------------------------
思路：

分成4个区间来考虑：

  if(A[mid] > first)  
     {  
       if( target >= first && target < A[mid])  
         end = mid;  
       else   
         start = mid;  
     }  
     else if( A[mid] < last)  
     {  
       if(target <= last && target > A[mid])  
         start = mid;  
       else  
        end =mid;  
     }  

实现：
极端情况是target恰好是first 或者 last，如果判断的时候不加等号会被漏掉。target 和mid比较的时候没必要加等号。