解题思路

本题中的子数组，称为递增子数组。
子数组指的是一个数组中一段连续的数组序列。
假设nums的长度为n，则nums的子数组的个数为$\frac{n\times (n+1)}{2}$。
因此只需要找出不满足的递增子数组即可。

python

class Solution:def incremovableSubarrayCount(self, nums: List[int]) -> int:n = len(nums)res = 0for i in range(n):for j in range(i, n):num = nums[:i] + nums[j+1:]if self.isIncrement(num):res +=1return resdef isIncrement(self, nums):for i in range(1,len(nums)):if nums[i] <= nums[i-1]:return Falsereturn True

实际上就是假设找到了移除递增子数组，则子数组的左边严格递增，右边也是严格递增，同样nums[l] > nums[r+1]。可以进一步简写得到：

class Solution:def incremovableSubarrayCount(self, nums: List[int]) -> int:n = len(nums)return sum([self.isIncreasing(nums[:i] + nums[j+1:]) for i in range(n) for j in range(i, n)])def isIncreasing(self, nums):# 判断是否严格递增的if len(nums) == 0:return Truereturn all(nums[i] < nums[i+1] for i in range(len(nums) - 1))

C++

c++不像python一样可以对列表直接截取，因此判断时传入i,j进行判断即可。

class Solution {
public:int incremovableSubarrayCount(vector<int>& nums) {int n = nums.size();int res = 0;for (int i = 0; i < n; i++){for (int j = i; j < n; j++){if (isIncreasing(nums, i, j)){res ++;}}}return res;}bool isIncreasing(vector<int>& nums, int l, int r){for (int i = 1; i < nums.size(); i++){// 处于l,r区间内if (i >=l && i <= r+1) continue;if (nums[i] <= nums[i-1]) return false;}// 再判断一下左右两边是否满足递增if (l-1 >= 0 && r+1 < nums.size() && nums[r+1] <=nums[l-1]) return false;return true;}
};

java

class Solution {public int incremovableSubarrayCount(int[] nums) {int n = nums.length;int res = 0;for (int i = 0; i < n; i++) {for (int j = i; j < n; j++) {if (isIncreasing(nums, i, j)) {res++;}}}return res;}public boolean isIncreasing(int[] nums, int l, int r) {for (int i = 1; i < nums.length; i++) {if (i >= l && i <= r + 1) {continue;} if (nums[i] <= nums[i - 1]) {return false;}}if (l - 1 >= 0 && r + 1 < nums.length && nums[r + 1] <= nums[l - 1]) {return false;}return true;}
}