给定一个整数数组 A,如果它是有效的山脉数组就返回 true,否则返回 false。
让我们回顾一下,如果 A 满足下述条件,那么它是一个山脉数组:
A.length >= 3
在 0 < i < A.length - 1 条件下,存在 i 使得:
A[0] < A[1] < ... A[i-1] < A[i]
A[i] > A[i+1] > ... > A[A.length - 1]
示例 1:
输入:[2,1]
输出:false
示例 2:
输入:[3,5,5]
输出:false
示例 3:
输入:[0,3,2,1]
输出:true
提示:
0 <= A.length <= 10000
0 <= A[i] <= 10000
本人愚笨的做法:时间复杂度最坏o(2n)
public class Shanmai {
public static boolean validMountainArray(int[] A) {
if(A.length < 3 || A[0] > A[1] || A[A.length-1] > A[A.length-2])
return false;
int pre = 0;
int aft = 0;
for (int i = 0; i < A.length-1; i++) {
if(A[i+1] == A[i]){
return false;
}
if(A[i+1]<A[i]){
pre = i;
break;
}
}
for (int i = A.length-1; i > 0; i--) {
if(A[i-1] == A[i]){
return false;
}
if(A[i]>A[i-1]){
aft = i;
break;
}
}
return pre == aft;
}
public static void main(String[] args) {
int a[] = {9,8,7,6,5,4,3,2,1,0};
System.out.println(validMountainArray(a));
}
}
大佬的:时间复杂度最坏 o(n)
public static boolean validMountainArray1(int[] A) {
if(A.length <3)
return false;
int left = 0;
int right = A.length-1;
while (left+1 <= right && A[left] >A[left+1]){
left++;
}
if (left == 0 || left == right){
return false;
}
while (left + 1 <= right && A[left] > A[left+1]){
left++;
}
return left == right;
}