描述

比如说，给你的数组是：

0, 2, 0, 4, 0, 1, 2

高度示意图：

  ^
  |
4 +        +--+
  |        |  |
3 +        |  |
  |        |  |
2 +  +--+  |  |     +--+
  |  |  |  |  |     |  |
1 +  |  |  |  |  +--+  |
  |  |  |  |  |  |  |  |
  +--+--+--+--+--+--+--+-->
   0  1  2  3  4  5  6

在这个高度图中，下标 2 位置的盛水量是 2，下标 4 位置的盛水量是 2，下标 5 位置的盛水量是 1，因此总共的盛水量是：

2 + 2 + 1 = 5

代码

public class AlgoCasts {

  // Time: O(n), Space: O(n)
  public int waterCanBeTrap(int[] height) {
    if (height == null || height.length == 0) return 0;
    int n = height.length, leftMax = -1, rightMax = -1, water = 0;
    int[] d = new int[n];
    for (int i = 0; i < n; ++i) {
      leftMax = Math.max(leftMax, height[i]);
      d[i] = leftMax;
    }
    for (int i = n-1; i >= 0; --i) {
      rightMax = Math.max(rightMax, height[i]);
      d[i] = Math.min(d[i], rightMax);
      water += (d[i] - height[i]);
    }
    return water;
  }

  // Time: O(n), Space: O(1)
  public int waterCanBeTrapO1(int[] height) {
    if (height == null || height.length == 0) return 0;
    int leftMax = -1, rightMax = -1, water = 0;
    int i = 0, j = height.length - 1;
    while (i <= j) {
      leftMax = Math.max(leftMax, height[i]);
      rightMax = Math.max(rightMax, height[j]);
      if (leftMax < rightMax) water += (leftMax - height[i++]);
      else water += (rightMax - height[j--]);
    }
    return water;
  }

}