Duplicate Zeros
往后加一个0，然后移动其他数 看到讨论里有个o(1) time, o(n) space的, 利用queue作为结果的预存储:
two pass
[LintCode] Product of Array Exclude Itself
public ArrayList<Long> productExcludeItself(ArrayList<Integer> A) { ArrayList<Long> res = new ArrayList<Long>(); if(A.size()==0 || A == null) return res; long p = 1; for(int i = 0 ; i < A.size(); i++) { res.add(p); p *= A.get(i); } p = A.get(A.size()-1); for(int i = A.size()-2; i >= 0; i–) { res.set(i,p*res.get(i)); p *= A.get(i); } return res; […]