Leetcode 42 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 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

這個題的意思是計算可以盛水的多少，可以使用雙指標，單次迴圈法，壓棧方法。

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）雙指標

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;
    }
}

注意細節，兩端的節點。