给你一个下标从 0 开始的整数数组 nums
。对于每个下标 i
(1 <= i <= nums.length - 2
),nums[i]
的 美丽值 等于:
2
,对于所有0 <= j < i
且i < k <= nums.length - 1
,满足nums[j] < nums[i] < nums[k]
1
,如果满足nums[i - 1] < nums[i] < nums[i + 1]
,且不满足前面的条件0
,如果上述条件全部不满足
返回符合 1 <= i <= nums.length - 2
的所有 nums[i]
的 美丽值的总和 。
示例 1:
输入:nums = [1,2,3] 输出:2 解释:对于每个符合范围 1 <= i <= 1 的下标 i : - nums[1] 的美丽值等于 2
示例 2:
输入:nums = [2,4,6,4] 输出:1 解释:对于每个符合范围 1 <= i <= 2 的下标 i : - nums[1] 的美丽值等于 1 - nums[2] 的美丽值等于 0
示例 3:
输入:nums = [3,2,1] 输出:0 解释:对于每个符合范围 1 <= i <= 1 的下标 i : - nums[1] 的美丽值等于 0
提示:
3 <= nums.length <= 105
1 <= nums[i] <= 105
C++
class Solution {
public:
int sumOfBeauties(vector<int>& nums) {
int n=nums.size();
vector<int> left(n,0);
vector<int> right(n,0);
int max_val=nums[0];
int min_val=nums[n-1];
for(int i=1;i<n-1;i++) {
if(nums[i]>max_val) {
left[i]=1;
}
max_val=max(max_val,nums[i]);
}
for(int i=n-2;i>0;i--) {
if(nums[i]<min_val) {
right[i]=1;
}
min_val=min(min_val,nums[i]);
}
int res=0;
for(int i=1;i<n-1;i++) {
if(left[i]==1 && right[i]==1) {
res+=2;
} else if(nums[i-1]<nums[i] && nums[i]<nums[i+1]) {
res+=1;
}
}
return res;
}
};
java
class Solution {
public int sumOfBeauties(int[] nums) {
int n = nums.length;
int[] left = new int[n];
int[] right = new int[n];
int max_val = nums[0];
for (int i = 1; i < n - 1; i++) {
if (nums[i] > max_val) {
left[i] = 1;
}
max_val = Math.max(max_val, nums[i]);
}
int min_val = nums[n - 1];
for (int i = n - 2; i > 0; i--) {
if (nums[i] < min_val) {
right[i] = 1;
}
min_val = Math.min(min_val, nums[i]);
}
int res = 0;
for (int i = 1; i < n - 1; i++) {
if (left[i] == 1 && right[i] == 1) {
res += 2;
} else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
res += 1;
}
}
return res;
}
}
python
class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
n = len(nums)
left = [0] * n
right = [0] * n
max_val = nums[0]
for i in range(1, n - 1):
if nums[i] > max_val:
left[i] = 1
max_val = max(max_val, nums[i])
min_val = nums[n - 1]
for i in range(n - 2, 0, -1):
if nums[i] < min_val:
right[i] = 1
min_val = min(min_val, nums[i])
res = 0
for i in range(1, n - 1):
if left[i] == 1 and right[i] == 1:
res += 2
elif nums[i - 1] < nums[i] and nums[i] < nums[i + 1]:
res += 1
return res