刷题笔记：对撞型／相会型指针（1） 灌水类

1 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.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

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!

» Solve this problem

[解题思路]
对于任何一个坐标，检查其左右的最大坐标，然后相减就是容积。所以，
1. 从左往右扫描一遍，对于每一个坐标，求取左边最大值。
2. 从右往左扫描一遍，对于每一个坐标，求最大右值。
3. 再扫描一遍，求取容积并加和。
#2和#3可以合并成一个循环，

 

扫描3遍数组的方法：

public class Solution {
    public int trap(int[] height) {
        if (height == null || height.length == 0) {
            return 0;
        }
        int len = height.length;
        int res = 0;
        int[] leftMax = new int[len];
        int[] rightMax = new int[len];
        leftMax[0] = height[0];
        rightMax[len - 1] = height[len - 1];
        for (int i = 1; i < len; i++) {
            leftMax[i] = Math.max(leftMax[i - 1], height[i]);
        }
        for (int i = len - 2; i >= 0; i--) {
            rightMax[i] = Math.max(rightMax[i + 1], height[i]);
        }
        for (int i = 1; i < len - 1; i++) {
            int amount = Math.min(leftMax[i - 1], rightMax[i + 1]) - height[i];
            if (amount > 0) {
                res += amount;
            }
        }
        return res;
    }
}

可以简化为对撞型指针，扫描1遍数组，总体思想类似

其实可以看作是一个动态规划的滚动数组优化，把leftMax, rightMax优化为o(1)变量

public class Solution {
    public int trap(int[] height) {
        if (height == null || height.length < 3) {
            return 0;
        }   
        int len = height.length;
        int leftHeight = height[0];
        int rightHeight = height[len - 1];
        int left = 0;
        int right = len - 1;
        int res = 0;
        
        while (left < right) {
            if (height[left] < height[right]) {
                if (leftHeight > height[left]) {
                    res += leftHeight - height[left];
                } else {
                    leftHeight = height[left];
                }
                left++;
            } else {
                if (rightHeight > height[right]) {
                    res+= rightHeight - height[right];
                } else {
                    rightHeight = height[right];
                }
                right--;
            }
        } // end of left < right
        return res;
    }
}

2. //todo