LeetCode 42 Trapping Rain Water (C++, Python)

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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 is able to trap after raining.
For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Solution C++


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class Solution {
public:
    int trap(vector<int>& height) {
        int ret = 0;
        if (height.empty()) return ret;
        vector<int> left(height.size(), 0);
        vector<int> right(height.size(), 0);
        int maxLeft = height[0];
        for (int i = 1; i < height.size(); ++i) {
            maxLeft = max(maxLeft, height[i - 1]);
            left[i] = maxLeft;
        }
        int maxRight = height[height.size() - 1];
        for (int i = height.size() - 2; i >= 0; --i) {
            maxRight = max(maxRight, height[i + 1]);
            right[i] = maxRight;
        }
        for (int i = 0; i < height.size(); ++i) {
            int tmp = min(left[i], right[i]) - height[i];
            if (tmp > 0) ret += tmp;
        }
        return ret;
    }
};
Solution Python


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class Solution:
    # @param height, an integer[]
    # @return an integer
    def trap(self, height):
        ret = 0
        if not height: return ret
        left = [0] * len(height)
        right = [0] * len(height)
        maxleft = height[0]
        for i in range(1, len(height)):
            maxleft = max(maxleft, height[i - 1])
            left[i] = maxleft
        maxright = height[-1]
        for i in range(len(height) - 2, -1, -1):
            maxright = max(maxright, height[i + 1])
            right[i] = maxright
        for i in range(len(height)):
            tmp = min(left[i], right[i]) - height[i]
            if tmp > 0:
                ret += tmp
        return ret

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