LeetCode 42 Trapping Rain Water

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本文介绍了一种计算降雨后地表积水体积的方法。通过寻找最高点并分别从两侧遍历，记录每个位置上的最大高度，当当前位置高度低于已知最大高度时，即形成积水区，其体积等于两者之差。

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!

题目链接：https://leetcode.com/problems/trapping-rain-water/

题目分析：考虑到存在高度差才会积水，所以先找最高的那个，然后分别从两边往最高的前进，并记录当前状态下的最高值，只要当前高度小于当前最高值，必然会积水，积水的面积就是高度差

public class Solution {
    public int trap(int[] height) {
        int maxHeight = 0, len = height.length, pos = 0;
        for(int i = 0; i < len; i ++) {
            if(maxHeight < height[i]) {
                pos = i;
                maxHeight = height[i];
            }
        }
        int ans = 0;
        int curMax = 0, curArea = 0;
        for(int i = 0; i < pos; i ++) {
            if(height[i] > curMax) {
                curMax = height[i];
            }
            else {
                curArea += curMax - height[i];
            }
        }
        ans += curArea;
        curMax = 0;
        curArea = 0;
        for(int i = len - 1; i > pos; i --) {
            if(height[i] > curMax) {
                curMax = height[i];
            }
            else {
                curArea += curMax - height[i];
            }
        }
        ans += curArea;
        return ans;
    }
}