Skip to content

Instantly share code, notes, and snippets.

Embed
What would you like to do?
Leetcode 54: Spiral matrix - clock wise - direction arrays, clockwise, and then range discussion
/*
Julia study the Leetcode discussion:
https://discuss.leetcode.com/topic/15558/a-concise-c-implementation-based-on-directions/2?page=1
What I like to do is to go over the notes and make it more readable first. Most of important is to train
myself to go over the thinking process closely. Spend more time on thinking process instead of the code.
When traversing the matrix in the spiral order, at any time we follow one out of the following four
directions: RIGHT DOWN LEFT UP. Suppose we are working on a 5 x 3 matrix as such:
0 1 2 3 4
6 7 8 9 10
11 12 13 14 15
Imagine a cursor starts off at (0, 0), i.e. the position at '0', then we can achieve
the spiral order by doing the following:
Go right 5 times
Go down 2 times
Go left 4 times
Go up 1 times.
Go right 3 times
Go down 0 times -> quit
**Range decrement by one for horizontal series**
**Rnage decerement by one for vertical series**
Notice that the directions we choose always follow the order 'right->down->left->up',
and for horizontal movements, the number of shifts follows: {5, 4, 3}, and vertical
movements follows {2, 1, 0}.
From the above example, we can define the horizontal range initial value is rows,
but the vertical initial value is cols - 1; next horizontal range will decrement value
by one, same as next vertical range. (Julia added this paragraph to help illustration.)
**One for loop instead of four**
We can make use of a direction matrix that records the offset for all directions, then
an array of two elements that stores the number of shifts for horizontal and vertical
movements, respectively. This way, we really just need one for loop instead of four.
Another good thing about this implementation is that: If later we decided to do spiral
traversal on a different direction (e.g. Counterclockwise), then we only need to change
the Direction matrix; the main loop does not need to be touched.
*/
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<vector<int> > dirs{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
vector<int> res;
int nr = matrix.size();
if (nr == 0) return res;
int nc = matrix[0].size();
if (nc == 0) return res;
vector<int> nSteps{nc, nr - 1};
int iDir = 0; // index of direction.
int ir = 0, ic = -1; // initial position
while (nSteps[iDir % 2])
{
for (int i = 0; i < nSteps[iDir%2]; ++i)
{
ir += dirs[iDir][0];
ic += dirs[iDir][1];
res.push_back(matrix[ir][ic]);
}
nSteps[iDir % 2]--;
iDir = (iDir + 1) % 4;
}
return res;
}
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment