[LeetCode]N-Queens

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https://leetcode.com/problems/n-queens/

回溯问题

做出来不难，但是空间复杂度有优化空间，我的做法是O(n2)，实际可以到O(n)

最优解：

保存三个cache数组，一个记录col已经被占用的位置，另外两个记录对角线被占用的位置。d1=rowCur+j d2=j-rowCur+n-1，d1、d2可以判断当前位置相对于对角线是否valid。

巧用API的重要性，大大简化代码量！！

public class Solution {    public List<List<String>> solveNQueens(int n) {        List<List<String>> res = new LinkedList<>();        if (n < 1) return res;        backTrace(res, new LinkedList<String>(), 0, n, new boolean[n], new boolean[n * 2], new boolean[n * 2]);        return res;    }    private void backTrace(List<List<String>> res, LinkedList<String> part, int curRow, int n, boolean[] col, boolean[] d1, boolean[] d2) {        if (curRow == n) {            res.add(new LinkedList<String>(part));            return;        }        for (int i = 0; i < n; i++) {            int t1 = i + curRow, t2 = n - i - 1 + curRow;            if (!col[i] && !d1[t1] && !d2[t2]) {                col[i] = true;                d1[t1] = true;                d2[t2] = true;                char[] arr = new char[n];                Arrays.fill(arr, '.');                arr[i] = 'Q';                part.add(new String(arr));                backTrace(res, part, curRow + 1, n, col, d1, d2);                col[i] = false;                d1[t1] = false;                d2[t2] = false;                part.remove(part.size() - 1);            }        }    }}

我的解法：

保存二维cache

public class Solution {    boolean[][] cache;    int n;    public List<List<String>> solveNQueens(int n) {        LinkedList<List<String>> res = new LinkedList<>();        if (n <= 0) return res;        cache = new boolean[n][n];        this.n = n;        for (int i = 0; i < n; i++) {            cache[0][i] = true;            dfs(1, res);            cache[0][i] = false;        }        return res;    }    private void dfs(int i, LinkedList<List<String>> res) {        if (i == n) {            LinkedList<String> temp = new LinkedList<>();            for (int j = 0; j < n; j++) {                StringBuilder sb = new StringBuilder();                for (int k = 0; k < n; k++) {                    if (cache[j][k]) {                        sb.append("Q");                    } else {                        sb.append(".");                    }                }                temp.add(sb.toString());            }            res.add(temp);            return;        }        for (int j = 0; j < n; j++) {            if (valid(i, j)) {                cache[i][j] = true;                dfs(i + 1, res);                cache[i][j] = false;            }        }    }    private boolean valid(int row, int col) {        for (int i = 0; i < row; i++) {            for (int j = 0; j < n; j++) {                if (cache[i][j] && (j == col || (Math.abs(i - row) == Math.abs(j - col)))) return false;            }        }        return true;    }}

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