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Turo店面,天哥manager,
先是BQ问一些项目问题,随后一道code问题,之后让我问问题,code如下:
/**
n = length of outer list 1 .. 10,000
m = max length of innner list
[[2,4,5], [45, 67], [2, 3]]
Output -
[2, 45, 2], [2, 67, 2], [4, 45, 2] .... ]
*
/
[2, 45, 2], [2, 45, 3], [2, 67, 2], [2, 67, 3], [4, 45, 2], [4, 45, 3], [4, 67, 2], [4, 67, 3] ....
for (i1) {
for (i2) {
.... return;
}
list.add(cur);
// main dfs
for (int next : inputs.get(curLevel)) {
dfs(inputs, rets, n, curLevl+1, next);
}
list.remove(list.size()-1); // backtracking
}
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