Leetcode trapping Rain Water

classSolution { Public:    intTrapintA[],intN) {if(N <1)return 0; Vector<int>Peaks; Vector<int>peaks_tmp; intLast_idx =0; BOOLincreasing =true;  for(intI=1; i<n; i++) {            if(A[i] <A[last_idx]) {                if(increasing) {peaks.push_back (LAST_IDX); } Increasing=false; } Else{Increasing=true; } last_idx=i; }        if(increasing) Peaks.push_back (N-1); if(Peaks.size () <2)return 0; BOOLUpdated =true;  while(updated) {Updated=false;            Peaks_tmp.clear (); Peaks_tmp.push_back (peaks[0]); Peaks_tmp.push_back (peaks[1]);  for(intI=2; I<peaks.size (); i++) {                intTlen =peaks_tmp.size (); intAi = Peaks_tmp[tlen-2]; intBI = Peaks_tmp[tlen-1]; intCI =Peaks[i]; if(A[ai] >= A[bi] && A[ci] >=A[bi]) {Peaks_tmp[tlen-1] =CI; Updated=true; } Else{peaks_tmp.push_back (CI);        }} swap (peaks, peaks_tmp); }        intRain =0;  for(intI=1; I<peaks.size (); i++) {            intleft = peaks[i-1]; intright=Peaks[i]; inth =min (A[left], a[right]); intBlocks =0;  for(intI=left +1; i<right; i++) blocks + = A[i] > h?H:a[i]; Rain+ = h * (Right-left-1) -blocks; }        returnRain; }};

The peaks of each block are calculated first, and then the rain must fall between the peak blocks, merging the bands bounded by the peak blocks (for sequence a{4,1,3,1,5}, the first interval is a (0,2), the second interval is a (2,4), because a[4] > a[2] and a[0] > a[2] So can be combined into intervals a (0, 4)), until the merger cannot be continued, ultimately calculating the amount of water that can accumulate in these block intervals.

A little bit better, the merging process takes place at peak time, and the extra array is reduced to a

classSolution { Public:        intTrapintA[],intN) {if(N <1)return 0; Vector<int>Peaks; intLast_idx =0; BOOLincreasing =true;  for(intI=1; i<n; i++) {            if(A[i] <A[last_idx]) {                if(increasing) {peaks.push_back (LAST_IDX);                Compact (A, peaks); } Increasing=false; } Else{Increasing=true; } last_idx=i; }                if(increasing) Peaks.push_back (N-1);                Compact (A, peaks); if(Peaks.size () <2)return 0; intRain =0;  for(intI=1; I<peaks.size (); i++) {            intleft = peaks[i-1]; intright=Peaks[i]; inth =min (A[left], a[right]); intBlocks =0;  for(intI=left +1; i<right; i++) blocks + = A[i] > h?H:a[i]; Rain+ = h * (Right-left-1) -blocks; }        returnRain; }    voidCompactintA[], vector<int> &peaks) {        intLen =peaks.size (); if(Len <3)return; BOOLUpdated =true;  while(Updated && (len = peaks.size ()) >=3) {Updated=false; intAi = Peaks[len-3]; intBI = Peaks[len-2]; intCI = Peaks[len-1]; if(A[ai] >= A[bi] && A[ci] >=A[bi]) {Peaks[len-2] =CI;                Peaks.pop_back (); Updated=true; }        }    }};

Another simple way to see: http://www.cnblogs.com/zhuli19901106/p/3542216.html, as follows

classSolution { Public:    intTrapintA[],intN) {if(N <2)return 0; Vector<int> Lmax (N,0); Vector<int> Rmax (N,0); intm = a[n-1];  for(inti=n-2; I >=0; i--) {            if(A[i] > m) m =A[i]; Rmax[i]=m; } m= a[0];  for(intI=1; i<n; i++) {            if(A[i] > m) m =A[i]; Lmax[i]=m; }        intRain =0;  for(intI=0; i<n; i++) {            inth =min (lmax[i], rmax[i]); intv = h-A[i]; Rain+ = v <0?0: v; }        returnRain; }};