[LeetCode]N-Queens

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
, 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
;
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

;
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;
}
}