力扣leetcode二分题33. 搜索旋转排序数组详细分析

Posted 2023-02-27 Afololer

题目

https://leetcode-cn.com/problems/search-in-rotated-sorted-array/submissions/

思考

往左边走还是往右边走？

猜想与mid与左右两端的关系有关，target与左右两端的关系有关

下面分情况讨论

mid指标的值比target要大

正常情况下，往左边寻找，然而经过了旋转，左边的小值可能旋转到右边

正常 [1,2,3,4,5,6,7]
旋转1 [2,3,4,5,6,7,1]
旋转2 [7,1,2,3,4,5,6]

显然，当mid比最右值小时，本次旋转为旋转2，只可能往左边走

那么假如mid比最右值大呢？

说明左边的小数旋转到了右边，此时mid左边的值必都比mid小,如果最左都比target大，那么只能往右边走，否则往左边走

mid指标的值比target要小

显然，当mid比最右值大时，本次旋转为旋转1，只可能往右边走

那么假如mid比最右值小呢？

根据以上分析，尝试情况都取反

说明右边的大数旋转到了左边，此时mid右边的值必都比mid大,如果最右都比target小，那么只能往左边走，否则往右边走

总结

if(nums[mid]<target)
	if nums[mid]<nums[high]
		high = mid-1;
	else 
		if nums[low]>target
		low = mid+1;
		else
		high = mid-1;
else
	if nums[mid]>nums[low]
		low = mid+1;
	else 
		if nums[high]<target
		high = mid-1;
		else
		low = mid+1;

代码

public int search(int[] nums, int target) 
        int len = nums.length;
        int low =0;
        int high = len-1;
        int mid;
        while(low<=high)
        
            mid = low+(high-low)/2;
            System.out.print(low);
            System.out.println(high);
            if(nums[mid]==target)
                return mid;
            if(nums[mid]>target)
                if(nums[mid]<nums[high])
                    high =mid-1;
                else
                    if(target<nums[low])
                        low = mid+1;
                    else
                        high=  mid-1;
                
            
            else
                 if(nums[mid]>nums[high])
                    low =mid+1;
                else
                    if(target>nums[high])
                        high=  mid-1;
                       
                    else
                         low = mid+1;
                
            

	       
        
        return -1;