LeetCode第51题 - N 皇后

题目

解答

class Solution {
    public List<List<String>> solveNQueens(int n) {
        int[][] chestPanel = new int[n][n];
        for (int i = 0; i < n; ++i) {
            Arrays.fill(chestPanel[i], 0);
        }
        List<List<String>> results = new ArrayList<>();

        solveNQueens(n, results, chestPanel, 0);

        return results;
    }

    public void solveNQueens(int n, List<List<String>> results, int[][] chestPanel, int row) {
        if (row == n) {
            List<String> lines = new ArrayList<>();
            for (int i = 0; i < n; ++i) {
                StringBuilder sb = new StringBuilder();
                for (int j = 0; j < n; ++j) {
                    if (chestPanel[i][j] == 1) {
                        sb.append("Q");
                    } else {
                        sb.append(".");
                    }
                }
                lines.add(sb.toString());
            }
            results.add(lines);
            return;
        }

        for (int i = 0; i < n; ++i) {
            if (isValidPosition(chestPanel, row, i)) {
                chestPanel[row][i] = 1;
                solveNQueens(n, results, chestPanel, row + 1);
                chestPanel[row][i] = 0;
            } else {
                chestPanel[row][i] = 0;
            }
        }
    }


    public boolean isValidPosition(int[][] chestPanel, int currentRow, int currentColomn) {

        int n = chestPanel.length;
        if (currentColomn >= n || currentRow >= n) {
            return false;
        }
        // 当前列
        for (int i = 0; i < currentRow; ++i) {
            if (chestPanel[i][currentColomn] == 1) {
                return false;
            }
        }

        // 左上角
        for (int i = currentRow - 1, j = currentColomn - 1; i >= 0 && j >= 0; --i, --j) {
            if (chestPanel[i][j] == 1) {
                return false;
            }
        }
        // 右上角
        for (int i = currentRow - 1, j = currentColomn + 1; i >= 0 && j < n; --i, ++j) {
            if (chestPanel[i][j] == 1) {
                return false;
            }
        }

        return true;
    }
}

总结

回溯法。