leetcode-153. Find Minimum in Rotated Sorted Array
2016-05-02 14:39
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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.
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思路:
1.当nums[0] < nums[nums.size()-1]时,未位移,直接返回nums[0]
2.否则说明位移了,已nums[0]为比较数进行折半,middle数字大于compare时继续后半部分,否则继续前半部分。注意nums[0]此时不能参与查找了,否者当middle == 0时可能会出问题
(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.
Subscribe to see which companies asked this question
思路:
1.当nums[0] < nums[nums.size()-1]时,未位移,直接返回nums[0]
2.否则说明位移了,已nums[0]为比较数进行折半,middle数字大于compare时继续后半部分,否则继续前半部分。注意nums[0]此时不能参与查找了,否者当middle == 0时可能会出问题
class Solution { public: int findMin(vector<int>& nums) { //两种情况,一种第一个比最后一个小,直接返回第一个 //另一种移位了,最小的在中间,折半和第一个元素比较 int length = nums.size(); if(length == 0) { return 0; } if(length == 1) { return nums[0]; } if(nums[0] < nums[length-1]) { return nums[0]; } //这要从下标1开始,nums[0]此时肯定不是最小的,不能参与排序,不然当最后middle==0时可能会出问题 return getMinElement(nums,1,length-1,nums[0]); } int getMinElement(vector<int>&nums,int begin,int end,int compareNum) { if(begin >= end) { return nums[begin]; } //防止溢出 int middle = begin + (end - begin)/2; if(nums[middle] > compareNum) { return getMinElement(nums,middle+1,end,compareNum); } else { //返回当前数字和另一半求得值得最小值 return min(nums[middle],getMinElement(nums,begin,middle-1,compareNum)); } } };
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