- , . for 3 - que, 5s . len , i for . , for , , len, , i 1 3.
x length(que)+1?
, = 5?
2 3 4 0 5 5 5 5 5 5? 2 3 4 5 5 5 5 5 5?
2 3 4 5 5 5 5 5 5 5? ( x)
matlab-ish @gnovice:
que = [2 3 4];
x = 5;
y = 10;
val = 5;
que(x:y)=val;
@gnovice:
que = [2 3 4];
x = 5;
y = 10;
val = 5;
que = [que, val.*ones(1,y-x+1)];
que = [que, repmat(val, 1, y-x+1)];
@woodchips:
que = [que, repmat(val, 1, y - length(que))];
In your situation, using a loop (for or at that time) is really bad, because in each loop you increase the size of the vector que with memory redistribution.
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