## 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

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。

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)
public class Solution {
public int trap(int[] A) {
final int n = A.length;
int[] left_peak = new int[n];
int[] right_peak = new int[n];

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

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

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)
public class Solution {
public int trap(int[] A) {
final int n = A.length;
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;
}
};