30 Day Challenge Day 18 | Leetcode 42. Trapping Rain Water

题解

双指针

三次遍历，但不影响复杂度为O(n)。后面要进行改进。

class Solution {
public:
    int trap(vector<int>& height) {
        vector<int> max_left(height.size()+2, 0);
        vector<int> max_right(height.size()+2, 0);
        
        for(int i = 1; i < max_left.size()-1; i++) {
            max_left[i] = max(max_left[i-1], height[i-1]);
        }

        for(int i = max_right.size()-2; i >= 1; i--) {
            max_right[i] = max(max_right[i+1], height[i-1]);
        }
        
        int max_water = 0;
        
        for(int i = 0; i < height.size(); i++) {
            cout << max_left[i+1] << ", " << max_right[i+1] << ", "<< height[i] << endl;
            max_water += max(min(max_left[i+1], max_right[i+1]) - height[i], 0);
        }
        
        return max_water;
    }
};

一次遍历，左右双指针。

class Solution {
public:
    int trap(vector<int>& height) {
        int water = 0;
        int l = 0, r = height.size()-1;
        while(l < r) {
            if(height[l] < height[r]) {
                int bar = height[l];
                l++;
                while(l < r && height[l] < bar) {
                    water += bar - height[l];
                    l++;
                }
            } else {
                int bar = height[r];
                r--;
                while(l < r && height[r] < bar) {
                    water += bar - height[r];
                    r--;
                }
            }
        }
        return water;
    }
};

方法：Stack

参考官方答案，可以用Stack来做这道题。我用题干中的例子，手动运行了每一步，才大致理解。比上述的方法是相对难掌握一点。

主要思想就是通过Stack来保存左边的Bar，只要发现有凹槽形状出现就开始更新水量。

class Solution {
public:
    int trap(vector<int>& height) {
        int water = 0, cur = 0;
        
        stack<int> st;
        
        while(cur < height.size()) {
            while(!st.empty() && height[cur] > height[st.top()]) {
                int top = st.top();
                st.pop();
                
                if(st.empty()) break; // no left bar
                
                int dist = cur - st.top() - 1;
                water += dist * (min(height[st.top()], height[cur]) - height[top]);
            }
            st.push(cur++);
        }
        
        return water;
    }
};