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.


题解:就是找数组中第一个“下凹”的地方,可以遍历,不过也可以用二分法,速度更快。

特别注意边界的处理:如果当前的中间值mid在0这个位置,只要考虑它后面的元素是不是小于它;如果在数组最后的位置上,要考虑它前面的元素是不是大于它。最后当发现当前mid所指的值不是所求的“下凹点”的时候,要把它和最后数组的最后一位比较,如果它比数组的最后一位大,说明要找的“下凹点”在mid所指位置的后面,否则在mid所指位置的前面。

JAVA版本代码如下:

 import java.awt.datatransfer.StringSelection;

 public class Solution {
public static void main(String[] args){
int num[] = {3,4,5,1,2};
Solution s = new Solution();
System.out.println(s.findMin(num));
}
public int findMin(int[] num) {
int begin = 0;
int end = num.length-1;
while(begin <= end){
int mid = (begin + end)/2;
if(mid == 0){
if(mid+1 < num.length && num[mid+1] < num[mid])
return num[mid+1];
else {
return num[0];
}
}
if(mid == num.length-1){
if(mid-1 >=0 && num[mid-1] > num[mid])
return num[mid];
else {
return num[0];
}
}
if(num[mid-1] > num[mid] && num[mid+1] > num[mid])
return num[mid]; if(num[num.length-1] < num[mid])
begin = mid+1;
else {
end = mid-1;
} }
return num[0];
}
}

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