# My easy-understanding Java solution with thinking process

• While, this is my first time solving a `hard` problem. After drawing on the paper, I get the idea:

• We need to find the V shapes to calculate the trapped water.
• Find the start of a decreasing sequence (p1)
• The end of the decreasing sequence (bottom)
• And the end of an increasing sequence (p2).
• V shape is bounded by p1, p2.
• I got WA since there might be nested V shapes.
• When calculating trapped water, fill the V shape.
• Repeat until the result remains the same.
• AC

Attached is my Java code, though it is looooong.

``````public class Solution {
public int trap(int[] height) {
int n = height.length;
if(n<3) return 0;
int p1=0,p2=1,bottom=0,i=0,j,count=0, count1=count;
while(true){
count1=count;
i=0;
while(i<n){
//find start of V shape
for(j=i;j<n;j++){
if(j==n-1) break;
if(height[j]>height[j+1]) break;
}
p1 = j;
//If p1 is the end of input, there is no bottom and end of the V shape.
if(p1==n-1) break;
//find bottom of V shape
for(j=p1+1;j<n-1;j++)
if(height[j]<height[j+1]) break;
bottom = j;
//find end of V shape
for(j=bottom+1;j<n-1;j++)
if(height[j]>height[j+1]) break;
p2 = j;
//Check whether it is a valid V shape: 0<=i<=bottom<=j<=n-1
if(p2>=n) break;
int tmp=height[p1]<height[p2]?height[p1]:height[p2];
for(j=p1+1;j<=p2-1;j++)
if(tmp>height[j]){
count+=(tmp-height[j]);
height[j]=tmp;
}
i=p2;
}
//If we all V shapes are calculated
if(count1==count) break;
}
return count;
}
}``````

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