数据范围:1 \le n \le 100001≤n≤10000,数组中任意元素的值: 0 \le val \le 100000≤val≤10000
要求:空间复杂度:O(1)O(1) ,时间复杂度:O(logn)O(logn)
import java.util.ArrayList; public class Solution { public int minNumberInRotateArray(int [] array) { int left = 0; int right = array.length - 1; while(left < right){ int mid = (left + right) / 2; //最小的数字在mid右边 if(array[mid] > array[right]) left = mid + 1; //无法判断,一个一个试 else if(array[mid] == array[right]) right--; //最小数字要么是mid要么在mid左边 else right = mid; } return array[left]; } }