LeetCode 42. Trapping Rain Water

题目链接

42. Trapping Rain Water

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. 

解：开一个水量数组，记录每个column当前水高，遍历时维持一个max和maxPos，分别是最高的column的高度和编号，如果当前 i 的column比前一个高，那么就更新 [maxPos,i] 的水量，代码：

class Solution {
public:
    int trap(vector<int>& height) {
    	int len = height.size();
    	if (len == 0) return 0;
    	int max = height[0];
    	int maxPos = 0;
    	int water = 0;
    	int *waterHeight = new int[len];
    	for (int i = 0; i < len; i++) waterHeight[i] = 0;
    	for (int i = 1; i < len; i++) {
    		if (height[i] > height[i-1]) {
    			if (height[i] >= max) {
    				for (int j = maxPos; j <= i; j++) waterHeight[j] = max;
    				maxPos = i;
    				max = height[i];
    			} else {
    				for (int j = maxPos; j <= i; j++) {
    					if (waterHeight[j] < height[i]) waterHeight[j] = height[i];
    				}
    			}
    		}
    	}
    	for (int i = 0; i < len; i++) {
    		if (height[i] < waterHeight[i]) water += waterHeight[i]-height[i];
    	} 
    	return water;
    }
};