Leetcode#52 N-Queens II

原题地址

跟N-Queens几乎没差别，好像还更简单了呢

代码：

int solve(int row, vector<bool> &col, vector<bool> left, vector<bool> right) {
  if (row == 0)
    return 1;

  vector<bool> avail(col.size(), false);
  for (int i = 0; i < col.size(); i++)
    avail[i] = !(col[i] || left[i] || right[i]);

  for (int i = 0; i < col.size() - 1; i++)
    left[i] = left[i + 1];
  left[col.size() - 1] = false;
  for (int i = col.size() - 1; i > 0; i--)
    right[i] = right[i - 1];
  right[0] = false;

  int res = 0;
  for (int i = 0; i < col.size(); i++) {
    if (!avail[i])
      continue;
    col[i] = true;
    if (i > 0) left[i - 1] = true;
    if (i < col.size() - 1) right[i + 1] = true;
    res += solve(row - 1, col, left, right);
    if (i < col.size() - 1) right[i + 1] = false;
    if (i > 0) left[i - 1] = false;
    col[i] = false;
  }

  return res;
}

int totalNQueens(int n) {
  vector<bool> col = vector<bool>(n, false);
  vector<bool> left = vector<bool>(n, false);
  vector<bool> right = vector<bool>(n, false);
  return solve(n, col, left, right);
}