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!

Subscribe to see which companies asked this question

Hide Tags

  Array Stack Two Pointers

Hide Similar Problems

  (M) Container With Most Water (M) Product of Array Except Self

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;
    }
}