Trapping Rain Water

描述

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.

分析

对于每个柱子，找到其左右两边最高的柱子，该柱子能容纳的面积就是min(max_left, max_right) - height。所以，

1. 从左往右扫描一遍，对于每个柱子，求取左边最大值；
2. 从右往左扫描一遍，对于每个柱子，求最大右值；
3. 再扫描一遍，把每个柱子的面积并累加。

也可以，

1. 扫描一遍，找到最高的柱子，这个柱子将数组分为两半；
2. 处理左边一半；
3. 处理右边一半。

代码1

// Trapping Rain Water
// 思路1，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
    int trap(const vector<int>& A) {
        const int n = A.size();
        int *left_peak = new int[n]();
        int *right_peak = new int[n]();

        for (int i = 1; i < n; i++) {
            left_peak[i] = max(left_peak[i - 1], A[i - 1]);
        }
        for (int i = n - 2; i >=0; --i) {
            right_peak[i] = max(right_peak[i+1], A[i+1]);
        }

        int sum = 0;
        for (int i = 0; i < n; i++) {
            int height = min(left_peak[i], right_peak[i]);
            if (height > A[i]) {
                sum += height - A[i];
            }
        }

        delete[] left_peak;
        delete[] right_peak;
        return sum;
    }
};

代码2

// Trapping Rain Water
// 思路2，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    int trap(const vector<int>& A) {
        const int n = A.size();
        int peak_index = 0; // 最高的柱子，将数组分为两半
        for (int i = 0; i < n; i++)
            if (A[i] > A[peak_index]) peak_index = i;

        int water = 0;
        for (int i = 0, left_peak = 0; i < peak_index; i++) {
            if (A[i] > left_peak) left_peak = A[i];
            else water += left_peak - A[i];
        }
        for (int i = n - 1, right_peak = 0; i > peak_index; i--) {
            if (A[i] > right_peak) right_peak = A[i];
            else water += right_peak - A[i];
        }
        return water;
    }
};

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