Trapping Rain Water

问题描述

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.

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!

 

解决思路

假设数组的长度为len，(1) 找到最高的那个柱子highest；

(2) 双指针：从0到highest, 从len-1到highest；

(3) 辅助栈s：存的是柱子高度的升序序列，如果遇到比栈顶元素小的高度，则能够储存水量为s.peek()-cur

时间空间复杂度均为O(n).

 

程序

public class Solution {
    public int trap(int[] height) {
		if (height == null || height.length == 0) {
			return 0;
		}
		int highest = getHighestIdx(height);
		int water = 0;
		
		Stack<Integer> s = new Stack<Integer>();
		for (int i = 0; i < highest; i++) {
			if (s.isEmpty() || height[i] > s.peek()) {
				s.push(height[i]);
			} else {
				water += s.peek() - height[i];
			}
		}
		
		s = new Stack<Integer>();
		for (int i = height.length - 1; i > highest; i--) {
			if (s.isEmpty() || height[i] > s.peek()) {
				s.push(height[i]);
			} else {
				water += s.peek() - height[i];
			}
		}
		
		return water;
	}
	
	private int getHighestIdx(int[] height) {
		int high = 0;
		int idx = 0;
		for (int i = 0; i < height.length; i++) {
			if (height[i] > high) {
				high = height[i];
				idx = i;
			}
		}
		return idx;
	}
}