目录
1. N 皇后
2. 搜索二维矩阵
3. 发奖金问题
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n 皇后问题 研究的是如何将 n
个皇后放置在 n×n
的棋盘上,并且使皇后彼此之间不能相互攻击。
给你一个整数 n
,返回所有不同的 n 皇后问题 的解决方案。
每一种解法包含一个不同的 n 皇后问题 的棋子放置方案,该方案中 'Q'
和 '.'
分别代表了皇后和空位。
示例 1:
输入:n = 4 输出:[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]] 解释:如上图所示,4 皇后问题存在两个不同的解法。
示例 2:
输入:n = 1 输出:[["Q"]]
提示:
1 <= n <= 9
代码:
import java.util.*;
public class solveNQueens {
public static class Solution {
public List> solveNQueens(int n) {
List> res = new ArrayList>();
int[] queenList = new int[n];
placeQueen(queenList, 0, n, res);
return res;
}
private void placeQueen(int[] queenList, int row, int n, List> res) {
if (row == n) {
ArrayList list = new ArrayList();
for (int i = 0; i < n; i++) {
String str = "";
for (int col = 0; col < n; col++) {
if (queenList[i] == col) {
str += "Q";
} else {
str += ".";
}
}
list.add(str);
}
res.add(list);
}
for (int col = 0; col < n; col++) {
if (isValid(queenList, row, col)) {
queenList[row] = col;
placeQueen(queenList, row + 1, n, res);
}
}
}
private boolean isValid(int[] queenList, int row, int col) {
for (int i = 0; i < row; i++) {
int pos = queenList[i];
if (pos == col) {
return false;
}
if (pos + row - i == col) {
return false;
}
if (pos - row + i == col) {
return false;
}
}
return true;
}
}
public static void main(String[] args) {
Solution s = new Solution();
System.out.println(s.solveNQueens(4));
}
}
输出:
[[.Q.., ...Q, Q..., ..Q.], [..Q., Q..., ...Q, .Q..]]
编写一个高效的算法来判断 m x n
矩阵中,是否存在一个目标值。该矩阵具有如下特性:
示例 1:
输入:matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3 输出:true
示例 2:
输入:matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13 输出:false
提示:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-10^4 <= matrix[i][j], target <= 10^4
代码:
import java.util.*;
public class searchMatrix {
public static class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0 || matrix[0].length == 0)
return false;
int begin, mid, end;
begin = mid = 0;
int len1 = matrix.length, len2 = matrix[0].length;
end = len1 * len2 - 1;
while (begin < end) {
mid = (begin + end) / 2;
if (matrix[mid / len2][mid % len2] < target)
begin = mid + 1;
else
end = mid;
}
return matrix[begin / len2][begin % len2] == target;
}
}
public static void main(String[] args) {
Solution s = new Solution();
int[][] matrix = {{1,3,5,7},{10,11,16,20},{23,30,34,60}};
System.out.println(s.searchMatrix(matrix, 3));
int[][] matrix2 = {{1,3,5,7},{10,11,16,20},{23,30,34,60}};
System.out.println(s.searchMatrix(matrix2, 13));
}
}
输出:
true
false
过年了,村里要庆祝。村长说,村里有一笔钱作为奖金,让每个人写一个纸条上来,谁写的数目与奖金最接近,就算猜中,这笔奖金就归谁,如果多人猜中,则平分。
编写程序,算算都有哪些人得到奖金?多少?
代码:
import java.util.Collections;
import java.util.Comparator;
import java.util.Arrays;
import java.io.PrintStream;
import java.nio.charset.StandardCharsets;
class Solution1 {
public static void main(String[] args) {
int award = 100;
String[] people = { "a", "b", "c", "d", "e", "f", "g", "h" };
Integer[] guess = { 75, 70, 80, 120, 100, 110, 100, 45 };
Integer[] ordered = new Integer[people.length];
for (int i = 0; i < ordered.length; i++)
ordered[i] = i;
Arrays.sort(ordered, new Comparator() {
@Override
public int compare(Integer a, Integer b) {
int x = guess[a] - award > 0 ? guess[a] - award : award - guess[a];
int y = guess[b] - award > 0 ? guess[b] - award : award - guess[b];
return x - y;
}
});
int maxp = 0;
int i = 0;
System.setOut(new PrintStream(System.out, true, StandardCharsets.UTF_8));
while (guess[ordered[i++]] == award)
maxp++;
if (maxp <= 1)
System.out.println(people[ordered[0]] + "一人得奖" + award + "元。");
else {
for (i = 0; i < maxp; i++)
System.out.print(people[ordered[i]] + " ");
System.out.println("共同得奖" + award / (float) (maxp) + "元。");
}
}
}
输出:
e g 共同得奖50.0元。
✨ 持续,努力奋斗做强刷题搬运工!
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