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.
在一个反转了的排好序的数组中,找出最小的数
可以直接寻找。
public class Solution { public int findMin(int[] nums) { if( nums.length == 1) return nums[0]; int result = nums[0]; for( int i = 1 ; i<nums.length;i++) result = Math.min(nums[i],result); return result; } }
二分查找。
public class Solution { public int findMin(int[] nums) { if( nums.length == 1) return nums[0]; int left = 0,right = nums.length-1; int mid = (left+right)/2; int first = nums[0]; while( left < right ){ if( nums[left] == nums[mid] ){ first = Math.min(Math.min(first,nums[left]),nums[right]); break; } else if( nums[left] < nums[mid] ){ left = mid; mid = (left+right)/2; first = Math.min(first,nums[left]); }else { right = mid; mid = (left+right)/2; first = Math.min(first,nums[left]); } } return Math.min(first,nums[left]); } }