feat: backtracking

Author: liguanliang1

leetcode/backtracking/index.js

@@ -0,0 +1,86 @@
+var permute = function (nums) {
+	let len = nums.length, res = [];
+	if (!len) return res;
+
+	let used = []; // boolean[]
+	let path = []; //number[]
+	dfs(nums, len, 0, path, used, res);
+	return res;
+
+	function dfs(nums, len, depth, path, used, res) {
+		if (depth === len) {
+			//path是动态数组，不能直接push，需要拷贝一份当前值保存到结果中
+			res.push([...path]);
+			return;
+		}
+
+		// 针对全排列中下标为 depth 的位置进行所有可能的尝试
+		for (let i = 0; i < len; i++) {
+			if (!used[i]) {
+				path.push(nums[i]);
+				used[i] = true;
+
+				console.log(path, '22222')
+				// 往下找全排列中的下一个位置
+				dfs(nums, len, depth + 1, path, used, res);
+
+				// 形成一个全排列后，进行回退，尝试其他答案
+				used[i] = false;
+				path.pop();
+			}
+		}
+	}
+};
+
+// 八皇后
+var solveNQueens = function (n) {
+	if (n == 0) return res;
+
+	let col = [], main = [], sub = []; // boolean[]
+	let res = []; // string[]
+	let path = []; //number[]
+	dfs(0, path);
+	return res;
+
+	function dfs(row, path) {
+		// 深度优先遍历到下标为 n，表示 [0.. n - 1] 已经填完，得到了一个结果
+		if (row == n) {
+			const board = convert2board(path);
+			res.push(board);
+			return;
+		}
+
+		// 针对下标为 row 的每一列，尝试是否可以放置
+		for (let j = 0; j < n; j++) {
+			if (!col[j] && !main[row - j + n - 1] && !sub[row + j]) {
+				path.push(j);
+
+				// 记录该位置的攻击范围
+				col[j] = true;
+				main[row - j + n - 1] = true; //加n-1是为了防止数组索引为负数
+				sub[row + j] = true;
+
+				// 进入下一行
+				dfs(row + 1, path);
+
+				// 回溯, 去掉path中最后一个值，尝试其他选项
+				col[j] = false;
+				main[row - j + n - 1] = false;
+				sub[row + j] = false;
+				path.pop();
+			}
+		}
+	}
+
+	// 输出一个结果
+	function convert2board(path) {
+		let board = []; // string[]
+		for (let i = 0; i < path.length; i++) {
+			let ret = new Array(n).fill('.');
+			ret[path[i]] = 'Q';
+			board.push(ret.join(''))
+		}
+		
+		return board;
+	}
+};

leetcode/dynamic-programing/water-problem.js

@@ -2,7 +2,7 @@ import LinkedList from '../link-list/LinkedList'
 
 
 // 积水问题，使用单调栈的思想
-function trap(height) {
+function trap0(height) {
 	const stack = new LinkedList();
 	// 最少也需要3个柱子才能积水，否则直接返回0
 	if (height.length < 3) return 0;
@@ -33,7 +33,7 @@ function trap(height) {
 	return res;
 }
 
-const result = trap([4, 2, 0, 3, 2, 5])
+const result = trap0([4, 2, 0, 3, 2, 5])
 console.log('water-problem', result)
 // // 按层计算，负责度较高可能会超时
 // class Solution42_byLayer {
@@ -147,7 +147,7 @@ console.log('water-problem', result)
 
 // 链接：https://leetcode-cn.com/problems/trapping-rain-water/solution/jie-yu-shui-by-leetcode/
 // 暴力解法
-function trap(height = []) {
+function trap1(height = []) {
 	let ans = 0;
 	let size = height.length;
 	for (let i = 1; i < size - 1; i++) {

leetcode/index.js

@@ -21,4 +21,5 @@ import "./stringCalculate"
 import "./MathSqrt"
 import "./search-binary-tree/index"
 import "./link-list/rotateList"
-import "./dynamic-programing/water-problem"
+import "./dynamic-programing/water-problem"
+import "./backtracking"