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* Update the book with the thrid revised edition * Fix a typo * Update the contributors' information * Update the mindmap * Update the version number
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@@ -4,7 +4,7 @@ | |
* Author: liuyuxin ([email protected]) | ||
*/ | ||
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/* 回溯算法:N 皇后 */ | ||
/* 回溯算法:n 皇后 */ | ||
void backtrack( | ||
int row, | ||
int n, | ||
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@@ -46,7 +46,7 @@ void backtrack( | |
} | ||
} | ||
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/* 求解 N 皇后 */ | ||
/* 求解 n 皇后 */ | ||
List<List<List<String>>> nQueens(int n) { | ||
// 初始化 n*n 大小的棋盘,其中 'Q' 代表皇后,'#' 代表空位 | ||
List<List<String>> state = List.generate(n, (index) => List.filled(n, "#")); | ||
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Original file line number | Diff line number | Diff line change |
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@@ -4,7 +4,7 @@ | |
* Author: Justin ([email protected]) | ||
*/ | ||
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/* 回溯算法:N 皇后 */ | ||
/* 回溯算法:n 皇后 */ | ||
function backtrack(row, n, state, res, cols, diags1, diags2) { | ||
// 当放置完所有行时,记录解 | ||
if (row === n) { | ||
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@@ -30,7 +30,7 @@ function backtrack(row, n, state, res, cols, diags1, diags2) { | |
} | ||
} | ||
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/* 求解 N 皇后 */ | ||
/* 求解 n 皇后 */ | ||
function nQueens(n) { | ||
// 初始化 n*n 大小的棋盘,其中 'Q' 代表皇后,'#' 代表空位 | ||
const state = Array.from({ length: n }, () => Array(n).fill('#')); | ||
|
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Original file line number | Diff line number | Diff line change |
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|
@@ -4,7 +4,7 @@ | |
* Author: codingonion ([email protected]) | ||
*/ | ||
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/* 回溯算法:N 皇后 */ | ||
/* 回溯算法:n 皇后 */ | ||
fn backtrack(row: usize, n: usize, state: &mut Vec<Vec<String>>, res: &mut Vec<Vec<Vec<String>>>, | ||
cols: &mut [bool], diags1: &mut [bool], diags2: &mut [bool]) { | ||
// 当放置完所有行时,记录解 | ||
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@@ -35,7 +35,7 @@ fn backtrack(row: usize, n: usize, state: &mut Vec<Vec<String>>, res: &mut Vec<V | |
} | ||
} | ||
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/* 求解 N 皇后 */ | ||
/* 求解 n 皇后 */ | ||
fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> { | ||
// 初始化 n*n 大小的棋盘,其中 'Q' 代表皇后,'#' 代表空位 | ||
let mut state: Vec<Vec<String>> = Vec::new(); | ||
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@@ -4,7 +4,7 @@ | |
* Author: nuomi1 ([email protected]) | ||
*/ | ||
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/* 回溯算法:N 皇后 */ | ||
/* 回溯算法:n 皇后 */ | ||
func backtrack(row: Int, n: Int, state: inout [[String]], res: inout [[[String]]], cols: inout [Bool], diags1: inout [Bool], diags2: inout [Bool]) { | ||
// 当放置完所有行时,记录解 | ||
if row == n { | ||
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@@ -34,7 +34,7 @@ func backtrack(row: Int, n: Int, state: inout [[String]], res: inout [[[String]] | |
} | ||
} | ||
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/* 求解 N 皇后 */ | ||
/* 求解 n 皇后 */ | ||
func nQueens(n: Int) -> [[[String]]] { | ||
// 初始化 n*n 大小的棋盘,其中 'Q' 代表皇后,'#' 代表空位 | ||
var state = Array(repeating: Array(repeating: "#", count: n), count: n) | ||
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Original file line number | Diff line number | Diff line change |
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@@ -4,7 +4,7 @@ | |
* Author: Justin ([email protected]) | ||
*/ | ||
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/* 回溯算法:N 皇后 */ | ||
/* 回溯算法:n 皇后 */ | ||
function backtrack( | ||
row: number, | ||
n: number, | ||
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@@ -38,7 +38,7 @@ function backtrack( | |
} | ||
} | ||
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/* 求解 N 皇后 */ | ||
/* 求解 n 皇后 */ | ||
function nQueens(n: number): string[][][] { | ||
// 初始化 n*n 大小的棋盘,其中 'Q' 代表皇后,'#' 代表空位 | ||
const state = Array.from({ length: n }, () => Array(n).fill('#')); | ||
|
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@@ -1,4 +1,4 @@ | ||
# N 皇后问题 | ||
# n 皇后问题 | ||
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!!! question | ||
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@@ -2,7 +2,7 @@ | |
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在本节中,我们先求解另一个常见的背包问题:完全背包,再了解它的一种特例:零钱兑换。 | ||
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## 完全背包 | ||
## 完全背包问题 | ||
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!!! question | ||
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Original file line number | Diff line number | Diff line change |
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@@ -1,4 +1,4 @@ | ||
# 图基础操作 | ||
# 图的基础操作 | ||
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图的基础操作可分为对“边”的操作和对“顶点”的操作。在“邻接矩阵”和“邻接表”两种表示方法下,实现方式有所不同。 | ||
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Binary file modified
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docs/chapter_preface/about_the_book.assets/hello_algo_mindmap.png
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