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]);
}
}

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