Java one pass solution, no extra space

• Well, my code looks a little longer because I have divided all the cases separately. It should be clear and easy to understand.

``````int biggercount;
int smallercount;
int lastpeak;
int total;
public int candy(int[] ratings) {
if(ratings.length == 0) return 0;
if(ratings.length == 1) return 1;
biggercount = 0; //used to record the number of children in the mono-increasing sequence
smallercount = 0;//used to record the number of children in the mono-decreasing sequence
lastpeak = 0;//record the number of candy needed of the highest requirement for last mono-increasing sequence
total = 0;
for(int i = 0;i<ratings.length;i++){
if(i==0) continue;
if(ratings[i]==ratings[i-1]){
if(smallercount!=0) countsmaller();
else if(biggercount!=0) countbigger();
else if(smallercount==0&&biggercount==0) total++;
lastpeak = 0;
}
else if(ratings[i]>ratings[i-1]){
if(smallercount!=0){
countsmaller();
total -= 1;//or the pit value would be calculated twice.
}
biggercount++;
}
else{
if(biggercount!=0) countbigger();
smallercount++;
}
if(i==ratings.length-1){
if(biggercount==0&&smallercount==0) total+=1;
else if(biggercount!=0) countbigger();
else if(smallercount!=0) countsmaller();
}
}
}
private void countsmaller(){
if(smallercount+1>lastpeak) total= total + calcusum(smallercount+1) - lastpeak;
else total += calcusum(smallercount);
smallercount = 0;
lastpeak = 0;
}
private void countbigger(){
total+=calcusum(biggercount+1);
lastpeak = biggercount + 1;
biggercount = 0;
}
private int calcusum(int k){//calculate arithmetic sequence
return (1+k)*k/2;
}``````

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