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TrappingRainWater(UsingArray).cpp
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TrappingRainWater(UsingArray).cpp
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/*
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
0 <= n <= 3 * 104
0 <= height[i] <= 105
*/
class Solution {
public:
int trap(vector<int>& height) {
if(height.size()==0){
return 0;
}
int *Left = new int [height.size()];
int *Right = new int [height.size()];
Left[0] = height[0];
Right[height.size()-1] = height[height.size()-1];
for(int i=1; i<height.size();i++){
Left[i] = max(Left[i-1],height[i]);
}
for(int i=height.size()-2; i>=0;i--){
Right[i] = max(Right[i+1],height[i]);
}
int ans = 0;
for(int i=0; i<height.size();i++){
ans += min(Left[i],Right[i])-height[i];
}
return ans;
}
};
/*
Runtime: 8 ms, faster than 85.32% of C++ online submissions for Trapping Rain Water.
Memory Usage: 14.7 MB, less than 28.06% of C++ online submissions for Trapping Rain Water.
*/