Trapping Rain Water *hard*

Given n non-negative integers representing an elevation map where the width of each bar are 1, compute how much WA ter It is the 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 the Rain Water (blue section) is being trapped.

intTrapintA[],intN) {intresult =0; //find the highest value/position    intMaxhigh =0; intMaxidx =0;  for(intI=0; i<n; i++){        if(A[i] >Maxhigh) {Maxhigh=A[i]; Maxidx=i; }    }    //From the left to the highest postion    intPrevhigh =0;  for(intI=0; i<maxidx; i++){        if(A[i] >Prevhigh) {Prevhigh=A[i]; } result+ = (Prevhigh-A[i]); }    //From the right to the highest postionPrevhigh=0;  for(inti=n-1; i>maxidx; i--){        if(A[i] >Prevhigh) {Prevhigh=A[i]; } result+ = (Prevhigh-A[i]); }    returnresult;}

* The idea is:
* 1) find the highest bar.
* 2) traverse the bar from left the highest bar.
* becasue we have the ' highest bar in right ' so ' any bar higher than it right bar (s) can contain the water.
* 3) traverse the bar from right the highest bar.
* becasue we have the highest bar in left, so, any bar higher than it left bar (s) can contain the water.

Trapping Rain Water *hard*