【LeetCode-153】Find Minimum in Rotated Sorted Array-优快云博客

本文链接：https://blog.youkuaiyun.com/cold2wind/article/details/50958433

本文介绍了两种在旋转排序数组中找到最小值的方法：一种是简单直观的O(n)复杂度遍历法；另一种则是更高效的O(log n)复杂度二分查找法，并提供了具体的Java代码实现。

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这道题刚开始看到,以为O(n)的复杂度肯定通不过,拖了两天,报着试试看的态度,直接把O(n)复杂度的代码粘了进去,这.....竟然通过了!!!

感觉还是太low了,后来用二分查找做了一下,代码如下

public class FindMinimuminRotatedSortedArray {
	
	//O(n)很好写
	public int findMin(int[] nums) {
        int min = nums[0];
        for(int i = 1;i < nums.length;i ++){
        	if(nums[i] < min){
        		min = nums[i];
        	}
        }
        
        return min;
    }
	
	//O(logn) 变种的二分查找法，判断条件变了
	public int findMinI(int[] nums) {
		int len = nums.length;  
        if (len == 1) 
        	return nums[0];  
        
        int left = 0, right = len - 1;  
        while (nums[left] > nums[right]) { // good idea  
            int mid = (left + right) / 2;  
            if (nums[mid] > nums[right]) {
                left = mid + 1;  
            } 
            else {  
                right = mid; // be careful, not mid-1, as num[mid] maybe the minimum  
            }  
        }  
        return nums[left];  
    }
	
	//O(logn)
	public int findMinII(int[] nums) {
        return findMinHelp(nums, 0, nums.length - 1);
    }
	
	private int findMinHelp(int[] nums,int begin,int end){
		if(nums[begin] <= nums[end]){
			return nums[begin];
		}
		
		int mid = begin + (end - begin) / 2;
		if(nums[mid] < nums[end]){
			return findMinHelp(nums, begin, mid);
		}
		//nums[mid] > nums[end] 则nums[mid]肯定不为最小值，所以mid + 1
		return findMinHelp(nums,mid + 1,end);
	}
}