-
Notifications
You must be signed in to change notification settings - Fork 0
/
0042. Trapping Rain Water
59 lines (50 loc) · 1.6 KB
/
0042. Trapping Rain Water
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
'''Given n non-negative integers representing an elevation map where the width of each bar is 1,
compute how much water it can trap after raining.
Example 1:
Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.
Example 2:
Input: height = [4,2,0,3,2,5]
Output: 9
Constraints:
n == height.length
1 <= n <= 2 * 104
0 <= height[i] <= 105
'''
class Solution:
def trap(self, height: List[int]) -> int:
l, r, res, = 0, len(height)-1, 0,
l_max, r_max = 0, 0
# l,r get meeting
while l < r:
if height[l] < height[r]:
if height[l] >= l_max: l_max = height[l]
else: res += l_max - height[l]
l+=1
else:
if height[r] >= r_max: r_max = height[r]
else: res += r_max - height[r]
r-=1
return res
class Solution {
public:
int trap(vector<int>& height)
{
int left = 0, right = height.size() - 1;
int ans = 0;
int left_max = 0, right_max = 0;
// l<r, ++l or --r until l=r
while (left < right) {
if (height[left] < height[right]) {
height[left] >= left_max ? (left_max = height[left]) : ans += (left_max - height[left]);
++left;
}
else {
height[right] >= right_max ? (right_max = height[right]) : ans += (right_max - height[right]);
--right;
}
}
return ans;
}
};