Trapping Rain Water解题报告

题目：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

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解题思路：从头到尾扫描数组，找到两个数，使得这两个数中间的所有数都小于等于这两个数，相当于是找到两个最高的矩形，然后计算这两个矩形中间的水的面积，然后叠加起来。Java代码：

public class Solution {
    public int trap(int[] height) {
        int sum = 0;
        int startPosition = 0;
        int index = 0;
        int endHeight = 0;
        int endPosition = 0;
        // 定位到第一个非0的位置
        while (startPosition < height.length && height[startPosition] == 0) {
            ++startPosition;
        }

        while (startPosition < height.length - 1) {
            index = startPosition + 1;
            endPosition = index;
            endHeight = height[endPosition];
            while (index < height.length) {
                if (height[startPosition] > height[index]) {
                    if (height[index] >= endHeight) {
                        endPosition = index;
                        endHeight = height[endPosition];
                    }
                    ++index;
                } else {
                    endPosition = index;
                    endHeight = height[endPosition];
                    break;
                }
            }
            sum += calculateTrapRain(height, startPosition, endPosition);
            startPosition = endPosition;
        }

        return sum;
    }

    private int calculateTrapRain(int[] height, int start, int end) {
        int lowerHeight = Math.min(height[start], height[end]);
        int sum = 0;
        for (int i = start + 1; i < end; ++i) {
            sum += (lowerHeight - height[i]);
        }
        return sum;
    }
   
    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] height = new int[] {0,1,0,2,1,0,1,3,2,1,2,1};
        System.out.println(solution.trap(height));
    }
}