LeetCode:Trapping Rain Water

Trapping Rain Water

Total Accepted: 68935 Total Submissions: 211305 Difficulty: Hard

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!

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思路一：

观察水的分布可发现，水位的分布情况显现梯形的，即先上升后下降。

因此可以先求出最高点，然后左右两边往中间遍历求水的容量。

c++ code:

class Solution {public:    int trap(vector<int>& height) {                int size = height.size();                int maxIndex = 0;        for(int i=0;i<size;i++) {            if(height[i] > height[maxIndex]) maxIndex = i;         }                int area = 0;        int maxLeft = 0;        for(int i=0;i<maxIndex;i++) {            if(height[i] > maxLeft) maxLeft = height[i];            else area += maxLeft - height[i];        }                int maxRight = 0;        for(int i=size-1;i>=maxIndex;i--) {            if(height[i] > maxRight) maxRight = height[i];            else area += maxRight - height[i];        }                return area;    }};

思路二：

直接用两个指针分别往中间移动，每次维持次高的水位secHeight。

java code:

public class Solution {    public int trap(int[] height) {                int len = height.length;                int left = 0, right = len - 1;        int secHeight = 0;        int area = 0;        while(left <right) {            if(height[left] < height[right]) {                secHeight = Math.max(secHeight, height[left]);                area += secHeight - height[left];                left++;            } else {                secHeight = Math.max(secHeight, height[right]);                area += secHeight - height[right];                right--;            }        }        return area;    }}