intfindMinVal(intmin,intmax,constvector<int> &rotateArray) {//取三个点的值intminVal = rotateArray[min];//最左边的点的值intmaxVal = rotateArray[max];//最右边的点的值intmidVal = rotateArray[(min + max) /2];//中间的点的值//终止条件if((max - min) <=1)returnminVal > maxVal ?max...
How and why? Suppose a sorted array 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. The array may contain duplicates. 解法1:本题旋转数组最小值的扩展,这次数组中可以存在重复元素了。上题因数组中...
Find the minimum element. The array may contain duplicates. Example 1: Input: [1,3,5] Output: 1 Example 2: Input: [2,2,2,0,1] Output: 0 Note: This is a follow up problem to Find Minimum in Rotated Sorted Array. Would allow duplicates affect the run-time complexity? How and why?
LeetCode: 154. Find Minimum in Rotated Sorted Array II 题目描述 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 ...
154. Find Minimum in Rotated Sorted Array II 题目 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 min element....
mid > start : 最小值在左半部分。 mid < start: 最小值在左半部分。 无论大于还是小于,最小值都在左半部分,所以 mid 和start 比较是不可取的。 mid 和end 比较 mid < end:最小值在左半部分。 mid > end:最小值在右半部分。 所以我们只需要把 mid 和end 比较,mid < end 丢弃右半部分(更新 en...
Can you solve this real interview question? Find Minimum in Rotated Sorted Array - Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become: * [4,5,6,7,0,1,2] if
Notice thatrotatingan array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]]. Given the sorted rotated array nums that may containduplicates, returnthe minimum element of this array. ...
由于是Rotated Sorted Array, 并且最低点在lo和hi之间,那么lo和hi之间只可能是 所以当noms[lo]<nums[hi]的时候,说明lo就是最低点。 当nums[mid]==nums[lo]的时候,并没有其它信息,只能将lo++, 因为lo和mid相等,可以缩短一个查找范围(mid仍在范围中) ...
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