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  • LeetCode | Find Minimum in Rotated Sorted Array

    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.

    tag : binary search

    //思路与前两天的题一样,即用middle的值与边缘值作比较,从而判断pivot是在middle的左或右
    //进而找到有序的一半数组,在这一半数组中找到最小的元素,再移动边缘
    public class Solution {
        public int findMin(int[] nums) {
            
            int result = Integer.MAX_VALUE;
            
            int left = 0;
            int right = nums.length - 1;
            
            while(left <= right){
                int middle = (left + right) / 2;
                
                if(nums[middle] < nums[right]){  //说明middle在pivot的右边,m-r段有序
                    if(nums[middle] < result){   //num[m]是m-r段最小的元素了
                        result = nums[middle];
                    }
                    right = middle - 1;
                }else{                           //题目指明无重复,此情况则l-m段有序
                    if(nums[left] < result){
                        result = nums[left];
                    }
                    left = middle + 1;          //关于while循环的终止,当循环到达left=middle时,
                }                               //无论如何下次会left>right从而终止循环
            }
            
            return result;
        }
    }


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  • 原文地址:https://www.cnblogs.com/dosmile/p/6444426.html
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