單詞搜索(079)

class Solution {
    int m, n;
    public boolean exist(char[][] board, String word) {
        m = board.length;
        n = board[0].length;
        char[] words = word.toCharArray();
        for(int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                if (backTrace(0, i, j, board, words)) return true;
            }
        }
        return false;
    }
    private boolean backTrace(int idx, int row, int col, char[][] board, char[] words){
        if (row < 0 || row >= m || col < 0 || col >= n || board[row][col] != words[idx]) return false;
        if (idx == words.length -1) return true;
        board[row][col] = '0';
        boolean status = backTrace(idx+1, row+1, col, board, words) || backTrace(idx+1, row, col+1, board, words)||
                         backTrace(idx+1, row-1, col, board, words) || backTrace(idx+1, row, col-1, board, words);
        board[row][col] = words[idx];
        return status;
    }
}

• 分析

簡單回溯, 開開胃

分割回文串(131)

class Solution {
    List<List<String>> res = new ArrayList<>();
    List<String> path = new ArrayList<>();
    public List<List<String>> partition(String s) {
        backTrace(0, s);
        return res;    
    }

    private void backTrace(int idx, String s){
        if (idx == s.length()){
            res.add(new ArrayList(path));
            return;
        }
        for (int j = idx; j < s.length(); j++){
            if (isPalindrome(idx, j, s)){
                path.add(s.substring(idx, j+1));
                backTrace(j+1, s);
                path.remove(path.size()-1);
            }
        }
    }

    private boolean isPalindrome(int lef, int rig, String s){
        while(lef < rig){
            if (s.charAt(lef++) != s.charAt(rig--)) return  false;
        }
        return true;
    }

}

• 分析

判斷是否為回文串, 若是則分割

N皇后(051)

class Solution {
    List<List<String>> res = new ArrayList<>();
    public List<List<String>> solveNQueens(int n) {
        int[] queens = new int[n];
        boolean[] column = new boolean[n];
        boolean[] attaRig = new boolean[2*n];
        boolean[] attaLef = new boolean[2*n];
        backTrace(0, queens, column, attaLef, attaRig);
        return res;
    }

    private void backTrace(int row, int[] queens, boolean[] column, boolean[] attaLef, boolean[] attaRig ){
        int n = column.length;
        if (row == n){
            List<String> temp = new ArrayList<>(n);
            for(int col : queens){
                char[] rowArray = new char[n];
                Arrays.fill(rowArray, '.');
                rowArray[col] = 'Q';
                temp.add(new String(rowArray));
            }
            res.add(temp);
            return;
        }

        for (int col = 0; col < n; col++){
            int attaLefIdx = row - col + n -1;
            if (!column[col] && !attaLef[attaLefIdx] && !attaRig[row + col] ){
                queens[row] = col;
                column[col] = attaLef[attaLefIdx] = attaRig[row + col] = true;
                backTrace(row+1, queens, column, attaLef, attaRig);
                column[col] = attaLef[attaLefIdx] = attaRig[row + col] = false;
            }
        }
    }
}

• 分析

N皇后帶來了一個條件 →<插入元素”行””列””對角線”不相容>

• 行不相容 →以行遍歷, 一行只插入一個元素
• 列, 對角線不相容 → Boolean數組來標記

靈神太強大了