Leetcode 42 Trapping Rain Water

最新推荐文章于 2021-01-04 11:29:16 发布

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最新推荐文章于 2021-01-04 11:29:16 发布

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本文链接：https://blog.youkuaiyun.com/Neo233/article/details/84393249

本文介绍了一种计算雨后地面积水的方法，通过分析地形高度数据，利用双指针、单次循环和压栈三种算法，计算出特定地形下雨水能够聚集的总量。示例中，对于给定的地形高度数组，算法能够准确计算出积水单位。

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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.

The above elevation map 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. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

这个题的意思是计算可以盛水的多少，可以使用双指针，单次循环法，压栈方法。

1）

class Solution {
    public int trap(int[] height) {
        if(height == null && height.length == 0){
            return 0;
        }
        int size = 0;
        int max = 0;
        int[] column = new int[height.length];
        
        for(int i = 0 ; i < height.length; ++i){
            column[i] = max;
            max = Math.max(max,height[i]);
        }
        max = 0;
        for(int i = height.length - 1 ; i >= 0; --i){
            column[i] = Math.min(max,column[i]);
            max = Math.max(max,height[i]);
            size += column[i] > height[i] ? Math.abs(column[i] - height[i]) : 0;
        }
        return size;
    }
}

2）双指针

class Solution {
    public int trap(int[] A) {
        int n = A.length;
        int left = 0, right = n - 1;
        int maxLeft = 0, maxRight = 0;
        int ans = 0;
        while (left < right) {
            if (A[left] < A[right]) {
                if (A[left] >= maxLeft) {
                    maxLeft = A[left];
                } else {
                    ans += maxLeft - A[left];
                }
                left ++;
            } else {
                if (A[right] >= maxRight) {
                    maxRight = A[right];
                } else {
                    ans += maxRight - A[right];
                }
                right --;
            }
        }
        return ans;
    }
}

3）

class Solution {
    //11.20
    public int trap(int[] height) {
        Stack<Integer> st = new Stack();
        int lastCounted = -1, lastCountedHeight, w,h;
        int area = 0;
        for (int i = 0; i< height.length; i++) {
            int curHeight = height[i];
            if (curHeight == 0)
                continue;
            
            while (!st.isEmpty()) {
                int lastBlock = st.peek();
              //  System.out.format("===%d %d %d %d %d %d\n", i, area, lastBlock,  height[lastBlock],curHeight,st.size());
                if (height[lastBlock] > curHeight) {
                    w = i - lastBlock - 1;
                    h = curHeight;//as curHeihgt is smaller
                    lastCountedHeight = 0;
                    if (lastCounted > lastBlock)
                        lastCountedHeight = height[lastCounted];
                    lastCounted = i;
                    area = area+ (w * h) - (w * lastCountedHeight);
                   // System.out.format("%d %d %d %d %d %d\n", i, area, w, h, lastCountedHeight,st.size());
                    st.push(i);
                    break;
                }
                st.pop();
                w = i - lastBlock - 1;
                h = height[lastBlock];
                lastCountedHeight = 0;
                if (lastCounted > lastBlock)
                    lastCountedHeight = height[lastCounted];
                lastCounted = lastBlock;
                area = area + (w * h) - (w * lastCountedHeight);
                
                //System.out.format("--%d %d %d %d %d  %d\n", i, area, w, h, lastCountedHeight, st.size());
            }
            //System.out.println(i);
            if (st.isEmpty()) {
                st.push(i);
            }
            
            
        }
        return area;
    }
}

注意细节，两端的节点。