LeetCode | Find Minimum in Rotated Sorted Array
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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.
You may assume no duplicate exists in the array.
这道题显然O(N)的解法是可解的,但是既然数组是有序的,其实可以采用O(lgN)复杂度内解决。
有一个十分十分重要的性质是:
a[left] < a[right]的时候,a[left]为最小值。
例如对于:
6 7 0 1 2 3 4 5
第一轮:left=0,right=7;mid=3,a[mid]=1;
a[left]>a[mid] right=mid=3;
第二轮:left=0,right=3,mid=1,a[mid]=7
a[left]< a[mid],left=mid+1=2;return 0;正确;
这是因为在a[left]>=a[right]的时候,必然会引起区间的收缩,且会向接近最大、最小值的方向收缩;
因为收缩的策略是a[left]>=a[right],这时候表示左侧有部分数据是被rotate过的,于是判断a[left]<=a[mid]否。
如果是,则最小值位于mid~right内,故left=mid+1;
本身二分没有太大难度,但是这个重要的性质还是值得好好体会。
class Solution {public: int findMin(vector<int>& nums) { int n=nums.size(); int left=0,right=n-1; while(left<right){ //非常重要的性质 if(nums[left]<nums[right]) return nums[left]; int mid=(left+right)/2; if(nums[left]<=nums[mid]) left=mid+1; else right=mid; } return nums[left]; }};
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