【发布时间】:2021-04-24 05:30:15
【问题描述】:
然而,我正在尝试学习反向跟踪,虽然我想将所有解决方案添加到我的数组列表中,但它只包含我的方法找到的第一个解决方案。我检查了 isSafe 方法,它是正确的。唯一的问题是我的queensList 方法。您能否解释一下如何将所有解决方案添加到我的数组列表中。 例如,对于 4x4 表,我的 ArrayList 的元素是那个
[0, 1, 0, 0]
[0, 0, 0, 1]
[1, 0, 0, 0]
[0, 0, 1, 0]
public static boolean queensList(int[][] aList, int r, int k, ArrayList<int[][]> qL) {
if (r >= k) {
qL.add(aList);
return true;
}
for (int i = 0; i < k; i++) {
if (isSafe(aList, r, i)) {
aList[r][i] = 1;
if (queensList(aList, r + 1, k, qL))
return true;
aList[r][i] = 0;
}
}
return false;
}
public static boolean isSafe(int[][] trying, int r, int c) {
for (int i = 0; i < trying.length; i++) {
if (trying[r][i] == 1 || trying[i][c] == 1)
return false;
}
int i = 0;
while (r - i < trying.length && 0 <= r - i && c - i < trying.length && 0 <= c - i || r - i < trying.length && 0 <= r - i && c + i < trying.length && 0 <= c + i
|| r + i < trying.length && 0 <= r + i && c - i < trying.length && 0 <= c - i || r + i < trying.length && 0 <= r + i && c + i < trying.length && 0 <= c + i) {
if (r - i < trying.length && 0 <= r - i && c - i < trying.length && 0 <= c - i && trying[r - i][c - i] == 1)
return false;
if (r - i < trying.length && 0 <= r - i && c + i < trying.length && 0 <= c + i && trying[r - i][c + i] == 1)
return false;
if (r + i < trying.length && 0 <= r + i && c - i < trying.length && 0 <= c - i && trying[r + i][c - i] == 1)
return false;
if (r + i < trying.length && 0 <= r + i && c + i < trying.length && 0 <= c + i && trying[r + i][c + i] == 1)
return false;
i++;
}
return true;
}
【问题讨论】:
标签: java backtracking n-queens recursive-backtracking