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Jz6.java
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89 lines (72 loc) · 1.96 KB
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package com.cpucode.java.simple;
/**
* 题目描述
* 把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。
* 输入一个非递减排序的数组的一个旋转，输出旋转数组的最小元素。
* NOTE：给出的所有元素都大于0，若数组大小为0，请返回0。
*
* 示例1
* 输入
* [3,4,5,1,2]
* 返回值
* 1
*
* @author : cpucode
* @Date : 2021/1/19
* @Time : 11:17
* @Github : https://github.com/CPU-Code
* @CSDN : https://blog.csdn.net/qq_44226094
*/
public class Jz6 {
public static void main(String[] args) {
Jz6 test = new Jz6();
int[] array = {3,4,5,1,2};
int num = test.minNumberInRotateArray(array);
System.out.println(num);
}
public int minNumberInRotateArray(int [] array) {
int num = 0;
//num = test1(array);
num = test2(array);
return num;
}
private int test1(int [] array){
if(array.length == 0){
return 0;
}
int num = array[0];
for(int a : array){
num = Math.min(num, a);
}
return num;
}
private int test2(int [] array){
/*数组的旋转形式 : [3,4,5,1,2]
[4,5,1,2,3]
*/
if(array.length == 0){
return 0;
}
int low = 0;
int high = array.length - 1;
while(low < high){
if(array[low] < array[high]){
//判断低位为最小值
return array[low];
}
//求中间位
int mid = (low + high) >> 1;
if(array[low] < array[mid]){
//说明为递增式
low = mid + 1;
}else if(array[mid] < array[high]){
//说明为递增式, 低值给高位
high = mid;
} else{
//无序
low ++;
}
}
return array[low];
}
}