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  • leetcode 二分查找 Search in Rotated Sorted ArrayII

    Search in Rotated Sorted Array II

     Total Accepted: 18500 Total Submissions: 59945My Submissions

    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.


    题意:一个已经排序好的数组。被按某个位置旋转了一次,给定一个值target,在该旋转后的数组里查找该值。(数组中的元素可能反复)
    思路:二分查找
    难点在于确定往数组的哪一半段继续二分查找
    设起点、中间点、终点分别为 start、middle、end (採用前闭后开的区间表示方法
    假设target = A[middle] return middle
    假设A[middle] > A[start],则[start,middle)单调递增
    1.假设target < A[middle] && target >= A[start],则 end = middle
    2.start = middle + 1, otherwise
    假设A[middle] < A[start],则[middle,end)单调递增
    1.假设target > A[middle] && target <= A[end - 1],则 start = middle + 1
    2.end = middle, otherwise
    假设A[middle] == A[start]。那A[start]也不会是target,能够通过start++; 去掉A[start]


    复杂度:时间O(n),空间O(1)


    int search(int A[], int n, int target){
    	int start = 0, end = n, middle ;
    	while(start < end){
    		middle = (start + end) / 2;
    		if(A[middle] == target) return middle;
    		if(A[middle] > A[start]){
    			if(target >= A[start] && target < A[middle]){
    				end = middle;
    			}else{
    				start = middle + 1;
    			}
    		}else if(A[middle] < A[start]){
    			if(target > A[middle] && target <= A[end - 1]){
    				start = middle + 1;
    			}else{
    				end = middle;
    			}
    		}
    		else{
    			++start;
    		}
    	}
    	return -1;
    }
    


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  • 原文地址:https://www.cnblogs.com/zhchoutai/p/6720072.html
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