CODE 91: Trapping Rain Water

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本文介绍了一种使用栈数据结构的方法来计算给定地形高度图中能够积存的雨水总量。通过遍历高度数组并利用栈记录当前高度及其索引，算法能够精确计算每个位置上的水容量。

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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!

	public int trap(int[] A) {
		// IMPORTANT: Please reset any member data you declared, as
		// the same Solution instance will be reused for each test case.
		if (A.length <= 2) {
			return 0;
		}
		Stack<Integer> hights = new Stack<Integer>();
		Stack<Integer> index = new Stack<Integer>();
		hights.push(A[0]);
		index.push(0);
		int sum = 0;
		for (int i = 1; i < A.length; i++) {
			if (A[i] > hights.peek()) {
				if (hights.size() <= 1) {
					hights.pop();
					index.pop();
					hights.push(A[i]);
					index.push(i);
				} else {
					while (A[i] > hights.peek() && hights.size() >= 2) {
						int before = hights.pop();
						int beforeIndex = index.pop();
						int bbefore = hights.pop();
						int bbeforeIndex = index.pop();
						int minBorad = Math.min(A[i], bbefore);
						sum += (i - bbeforeIndex - 1) * (minBorad - before);
						hights.push(bbefore);
						index.push(bbeforeIndex);
					}
					if (hights.size() <= 1 && A[i] > hights.peek()) {
						hights.pop();
						index.pop();
						hights.push(A[i]);
						index.push(i);
					} else if (A[i] < hights.peek()) {
						hights.push(A[i]);
						index.push(i);
					} else if (A[i] == hights.peek()) {
						index.pop();
						index.push(i);
					}
				}
			} else if (A[i] < hights.peek()) {
				hights.push(A[i]);
				index.push(i);
			} else {
				index.pop();
				index.push(i);
			}
		}
		return sum;
	}