时间复杂度为O(N)的很好达到,这里主要是列出二分查找的方法,时间复杂度是O(logN)
numbers[mid] == numbers[r]时,r = r - 1是关键。
class Solution {
public int minArray(int[] numbers) {
if(numbers.length < 1){
return 0;
}
/*for(int i = 0;i < numbers.length-1;i++){
if(numbers[i] > numbers[i+1]){
return numbers[i+1];
}
}
return numbers[0];*/
int l = 0,r = numbers.length-1;
while(l < r){
int mid = l + r >> 1;
if(numbers[mid] > numbers[r]){
l = mid+1;
}else if(numbers[mid] < numbers[r]){
r = mid;
}else{
r = r - 1;//缩小边界
}
}
return numbers[l];
}
}