題目：Find Minimum in Rotated Sorted Array

難度：medium

問題描述：

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]).

Find the minimum element.

You may assume no duplicate exists in the array.

Example 1:

Input: [3,4,5,1,2] 
Output:
 
 1

Example 2:

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

解題思路：

兩點法即可，不過要考慮沒有摺疊時的特殊情況。154在於可能有重複元素。

程式碼如下：

public static int findMin(int[] nums) {
        int len = nums.length;
        if(nums.length == 1) return nums[0];
        int p1 = 0,p2 = len-1;
        if(nums[len-1]>nums[0]) return nums[0];
        while(true){
        	if(p1>=p2-1) return Math.min(nums[p1], nums[p2]);
        	int mid = (p1+p2)>>1;
        	//System.out.println(mid);
        	if(nums[mid]>=nums[p1]){	//mid在前半段
        		p1 = mid;
        		continue;
        	}else{	//mid在後半段
        		p2 = mid;
        		continue;
        	}
        }
    }
 

154程式碼如下：

public int findMin2(int[] nums) {
		int len = nums.length;
        if(nums.length == 1) return nums[0];
        int p1 = 0,p2 = len-1,mid = 0;
        if(nums[len-1]>nums[0]) return nums[0];
        while(p1<p2){
        	//3,4,5,1,2	--情形1
        	//1,2,3,4,5 --情形2
        	mid = (p1+p2)>>1;
        	if(nums[mid]>nums[p2]){//只存在於情形1
        		p1 = mid+1;
        	}else if(nums[mid]<nums[p2]){//情形1和2都有
        		p2 = mid;
        	}else{//nums[mid]==nums[p2],p2即使是最小值也不影響
        		p2 = p2-1;
        	}
        }
        return nums[p1];
    }