Trapping Rain Water
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.
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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
Solution
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class Solution {
public:
int trap(vector<int>& height) {
int solidArea = 0;
int topIndex = 0;
int topHeight = 0;
int width = height.size();
for (int i = 0; i < width; i++) {
solidArea += height[i];
if (topHeight < height[i]) {
topIndex = i;
topHeight = height[i];
}
}
for (int left = 0; left < topIndex; left++) {
if (height[left + 1] < height[left]) {
height[left + 1] = height[left];
}
}
for (int right = width - 1; right > topIndex; right--) {
if (height[right] > height[right - 1]) {
height[right - 1] = height[right];
}
}
int fixedArea = 0;
for (auto item : height) {
fixedArea += item;
}
return fixedArea - solidArea;
}
};
