[抄题]:
An array is monotonic if it is either monotone increasing or monotone decreasing.
An array A
is monotone increasing if for all i <= j
, A[i] <= A[j]
. An array A
is monotone decreasing if for all i <= j
, A[i] >= A[j]
.
Return true
if and only if the given array A
is monotonic.
Example 1:
Input: [1,2,2,3]
Output: true
Example 2:
Input: [6,5,4,4]
Output: true
Example 3:
Input: [1,3,2]
Output: false
Example 4:
Input: [1,2,4,5]
Output: true
Example 5:
Input: [1,1,1]
Output: true
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
想到了用inc dec变量来记录,以为要判断第一个:不用,两边同时加就行了
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
自己跑Case很重要
[复杂度]:Time complexity: O(N) Space complexity: O(1)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
class Solution { public boolean isMonotonic(int[] A) { //corner case if (A == null || A.length == 0) return true; //for loop int inc = 0; int dec = 0; /* 1 2 2 3 inc 1 2 3 dec 1 */ for (int i = 1; i < A.length; i++) { if (A[i] > A[i -1]) inc++; else if (A[i] < A[i -1]) dec++; else { inc++; dec++; } } //return return (inc == A.length - 1) || (dec == A.length - 1); } }