LeetCode OJ 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. Thanks Marcos for contributing this image!
【题目分析】
题目给出一个整数序列代表高度图,要求计算下雨后图中的积水量。
【思路】
我们只要找到高度图中的凹陷部分,然后把所有凹陷的容积加起来即可。那么如何找到这些凹陷部分呢?
有这样一个想法:给定高度图两端的高度,那些比这两端都低的部分肯定会形成凹陷,如果是相等则不形成凹陷,如果比当前高度高,则我们要重新确定高度图的两端高度。这个过程如下:
1. 初始两端最低高度high = 0, 容量capacity = 0;
2. height[begin] <= high,不形成凹陷,begin++;
3. height[begin] > high; height[end] > high;high = min(height[begin], height[end]); 因此high = 1;
此时height[begin] == high,height[end] == high;所以begin++,end--;
4. height[begin] < high;形成凹陷,capacity += high - height[begin];此时capacity = 1;begin++;
5. height[begin] > high; height[end] > high;high = min(height[begin], height[end]); 因此high = 2;
此时height[begin] == high,height[end] == high;所以begin++,end--;
6. height[begin] < high;形成凹陷,capacity += high - height[begin];此时capacity = 2;begin++;
height[end] < high;形成凹陷,capacity += high - height[begin];此时capacity = 3;end--;
7. height[begin] < high;形成凹陷,capacity += high - height[begin];此时capacity = 5;begin++;
height[end] = high;不形成凹陷;end--;
7. height[begin] < high;形成凹陷,capacity += high - height[begin];此时capacity = 6;begin++;
8. 返回最大容量6;
【java代码】
public class Solution {
public int trap(int[] height) {
if(height == null || height.length <= 2) return 0; int sum = 0, maxhigh = 0;
int begin = 0, end = height.length-1;
while(begin <= end){
if(height[begin] < maxhigh){
sum += maxhigh - height[begin++];
}
else if(height[end] < maxhigh){
sum += maxhigh - height[end--];
}
else{
maxhigh = Math.min(height[begin], height[end]);
if(height[begin] <= maxhigh) begin++;
if(height[end] <= maxhigh) end--;
}
}
return sum;
}
}
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