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  • 154. Find Minimum in Rotated Sorted Array II (Array; Divide-and-Conquer)

    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 element.

    The array may contain duplicates.

    思路:有重复元素的时候,不能按I中这么分三类(反例:Input: [3,3,1,3]=>将roate in right判断成没有rotate)。解决方法是,当碰到nums[start]=nums[end]的情况时,end-1,寻找不同元素再进行二分法。

    class Solution {
    public:
        int findMin(vector<int>& nums) {
            return dfs(nums,0,nums.size()-1);
        }
        
        int dfs(vector<int>& nums, int start, int end){
            if(start==end) return nums[start];
            if(nums[start]==nums[end]) return dfs(nums,start, end-1);
            
            int mid = start + ((end - start) >> 1);
            if(nums[mid] > nums[end]){ //rotate in the right
                return dfs(nums, mid+1,end);
            }
            else if(nums[start] <= nums[mid]){ //no rotate
                return nums[start];
            }
            else{ //rorate in the left
                return dfs(nums, start, mid);
            }
        }
    };
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  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/6291545.html
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