From 90540c8bb8257dc45be56a27e21f3006df05450a Mon Sep 17 00:00:00 2001 From: CyC2018 <1029579233@qq.com> Date: Wed, 28 Mar 2018 16:03:28 +0800 Subject: [PATCH] auto commit --- notes/Leetcode 题解.md | 287 +++++++++++++++++++++-------------------- 1 file changed, 147 insertions(+), 140 deletions(-) diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md index 034533d7..73faaea7 100644 --- a/notes/Leetcode 题解.md +++ b/notes/Leetcode 题解.md @@ -13,14 +13,14 @@ * [Backtracking](#backtracking) * [分治](#分治) * [动态规划](#动态规划) - * [分割整数](#分割整数) - * [矩阵路径](#矩阵路径) * [斐波那契数列](#斐波那契数列) * [最长递增子序列](#最长递增子序列) * [最长公共子系列](#最长公共子系列) * [0-1 背包](#0-1-背包) * [数组区间](#数组区间) * [字符串编辑](#字符串编辑) + * [分割整数](#分割整数) + * [矩阵路径](#矩阵路径) * [其它问题](#其它问题) * [数学](#数学) * [素数](#素数) @@ -1719,130 +1719,6 @@ public List diffWaysToCompute(String input) { 递归和动态规划都是将原问题拆成多个子问题然后求解，他们之间最本质的区别是，动态规划保存了子问题的解。 -### 分割整数 - -**分割整数的最大乘积** - -[Leetcode : 343. Integer Break (Medim)](https://leetcode.com/problems/integer-break/description/) - -题目描述：For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4). - -```java -public int integerBreak(int n) { - int[] dp = new int[n + 1]; - dp[1] = 1; - for(int i = 2; i <= n; i++) { - for(int j = 1; j <= i - 1; j++) { - dp[i] = Math.max(dp[i], Math.max(j * dp[i - j], j * (i - j))); - } - } - return dp[n]; -} -``` - -**按平方数来分割整数** - -[Leetcode : 279. Perfect Squares(Medium)](https://leetcode.com/problems/perfect-squares/description/) - -题目描述：For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9. - -```java -public int numSquares(int n) { - List squareList = generateSquareList(n); - int[] dp = new int[n + 1]; - for (int i = 1; i <= n; i++) { - int max = Integer.MAX_VALUE; - for (int square : squareList) { - if (square > i) break; - max = Math.min(max, dp[i - square] + 1); - } - dp[i] = max; - } - return dp[n]; -} - -private List generateSquareList(int n) { - List squareList = new ArrayList<>(); - int diff = 3; - int square = 1; - while (square <= n) { - squareList.add(square); - square += diff; - diff += 2; - } - return squareList; -} -``` - -**分割整数构成字母字符串** - -[Leetcode : 91. Decode Ways (Medium)](https://leetcode.com/problems/decode-ways/description/) - -题目描述：Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12). - -```java -public int numDecodings(String s) { - if(s == null || s.length() == 0) return 0; - int n = s.length(); - int[] dp = new int[n + 1]; - dp[0] = 1; - dp[1] = s.charAt(0) == '0' ? 0 : 1; - for(int i = 2; i <= n; i++) { - int one = Integer.valueOf(s.substring(i - 1, i)); - if(one != 0) dp[i] += dp[i - 1]; - if(s.charAt(i - 2) == '0') continue; - int two = Integer.valueOf(s.substring(i - 2, i)); - if(two <= 26) dp[i] += dp[i - 2]; - } - return dp[n]; -} -``` - -### 矩阵路径 - -**矩阵的总路径数** - -[Leetcode : 62. Unique Paths (Medium)](https://leetcode.com/problems/unique-paths/description/) - -题目描述：统计从矩阵左上角到右下角的路径总数，每次只能向右或者向下移动。 - -

- -```java -public int uniquePaths(int m, int n) { - int[] dp = new int[n]; - for (int i = 0; i < m; i++) { - for (int j = 0; j < n; j++) { - if(i == 0) dp[j] = 1; - else if(j != 0) dp[j] = dp[j] + dp[j - 1]; - } - } - return dp[n - 1]; -} -``` - -**矩阵的最小路径和** - -[Leetcode : 64. Minimum Path Sum (Medium)](https://leetcode.com/problems/minimum-path-sum/description/) - -题目描述：求从矩阵的左上角到右下角的最小路径和，每次只能向左和向下移动。 - -```java -public int minPathSum(int[][] grid) { - if(grid.length == 0 || grid[0].length == 0) return 0; - int m = grid.length, n = grid[0].length; - int[] dp = new int[n]; - for(int i = 0; i < m; i++) { - for(int j = 0; j < n; j++) { - if(j == 0) dp[0] = dp[0] + grid[i][0]; - else if(i == 0) dp[j] = dp[j - 1] + grid[0][j]; - else dp[j] = Math.min(dp[j - 1], dp[j]) + grid[i][j]; - } - } - return dp[n - 1]; -} -``` - ### 斐波那契数列 **爬楼梯** @@ -1863,7 +1739,6 @@ dp[N] 即为所求。 public int climbStairs(int n) { if(n == 1) return 1; if(n == 2) return 2; - // 前一个楼梯、后一个楼梯 int pre1 = 2, pre2 = 1; for(int i = 2; i < n; i++){ int cur = pre1 + pre2; @@ -1937,31 +1812,32 @@ public int rob(int[] nums) { [Leetcode : 213. House Robber II (Medium)](https://leetcode.com/problems/house-robber-ii/description/) ```java +private int[] dp; + public int rob(int[] nums) { - if(nums == null || nums.length == 0) return 0; + if (nums == null || nums.length == 0) return 0; int n = nums.length; - if(n == 1) return nums[0]; + if (n == 1) return nums[0]; + dp = new int[n]; return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1)); } -private int rob(int[] nums, int s, int e) { - int n = nums.length; - if(e - s == 0) return nums[s]; - if(e - s == 1) return Math.max(nums[s], nums[s + 1]); - int[] dp = new int[n]; - dp[s] = nums[s]; - dp[s + 1] = nums[s + 1]; - dp[s + 2] = nums[s] + nums[s + 2]; - for (int i = s + 3; i <= e; i++) { +private int rob(int[] nums, int first, int last) { + if (last - first == 0) return nums[first]; + if (last - first == 1) return Math.max(nums[first], nums[first + 1]); + dp[first] = nums[first]; + dp[first + 1] = nums[first + 1]; + dp[first + 2] = nums[first] + nums[first + 2]; + for (int i = first + 3; i <= last; i++) { dp[i] = Math.max(dp[i - 2], dp[i - 3]) + nums[i]; } - return Math.max(dp[e], dp[e - 1]); + return Math.max(dp[last], dp[last - 1]); } ``` **信件错排** -题目描述：有 N 个 信 和 信封，它们被打乱，求错误装信的方式数量。 +题目描述：有 N 个 信 和 信封，它们被打乱，求错误装信方式的数量。 定义一个数组 dp 存储错误方式数量，dp[i] 表示前 i 个信和信封的错误方式数量。假设第 i 个信装到第 j 个信封里面，而第 j 个信装到第 k 个信封里面。根据 i 和 k 是否相等，有两种情况： @@ -2545,6 +2421,137 @@ public int minDistance(String word1, String word2) { [Leetcode : 72. Edit Distance (Hard)](https://leetcode.com/problems/edit-distance/description/) +### 分割整数 + +**分割整数的最大乘积** + +[Leetcode : 343. Integer Break (Medim)](https://leetcode.com/problems/integer-break/description/) + +题目描述：For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4). + +```java +public int integerBreak(int n) { + int[] dp = new int[n + 1]; + dp[1] = 1; + for(int i = 2; i <= n; i++) { + for(int j = 1; j <= i - 1; j++) { + dp[i] = Math.max(dp[i], Math.max(j * dp[i - j], j * (i - j))); + } + } + return dp[n]; +} +``` + +**按平方数来分割整数** + +[Leetcode : 279. Perfect Squares(Medium)](https://leetcode.com/problems/perfect-squares/description/) + +题目描述：For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9. + +```java +public int numSquares(int n) { + List squareList = generateSquareList(n); + int[] dp = new int[n + 1]; + for (int i = 1; i <= n; i++) { + int max = Integer.MAX_VALUE; + for (int square : squareList) { + if (square > i) break; + max = Math.min(max, dp[i - square] + 1); + } + dp[i] = max; + } + return dp[n]; +} + +private List generateSquareList(int n) { + List squareList = new ArrayList<>(); + int diff = 3; + int square = 1; + while (square <= n) { + squareList.add(square); + square += diff; + diff += 2; + } + return squareList; +} +``` + +**分割整数构成字母字符串** + +[Leetcode : 91. Decode Ways (Medium)](https://leetcode.com/problems/decode-ways/description/) + +题目描述：Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12). + +```java +public int numDecodings(String s) { + if(s == null || s.length() == 0) return 0; + int n = s.length(); + int[] dp = new int[n + 1]; + dp[0] = 1; + dp[1] = s.charAt(0) == '0' ? 0 : 1; + for(int i = 2; i <= n; i++) { + int one = Integer.valueOf(s.substring(i - 1, i)); + if(one != 0) dp[i] += dp[i - 1]; + if(s.charAt(i - 2) == '0') continue; + int two = Integer.valueOf(s.substring(i - 2, i)); + if(two <= 26) dp[i] += dp[i - 2]; + } + return dp[n]; +} +``` + +### 矩阵路径 + +**矩阵的总路径数** + +[Leetcode : 62. Unique Paths (Medium)](https://leetcode.com/problems/unique-paths/description/) + +题目描述：统计从矩阵左上角到右下角的路径总数，每次只能向右或者向下移动。 + +

+ +```java +public int uniquePaths(int m, int n) { + int[] dp = new int[n]; + Arrays.fill(dp, 1); + for (int i = 1; i < m; i++) { + for (int j = 1; j < n; j++) { + dp[j] = dp[j] + dp[j - 1]; + } + } + return dp[n - 1]; +} +``` + +**矩阵的最小路径和** + +[Leetcode : 64. Minimum Path Sum (Medium)](https://leetcode.com/problems/minimum-path-sum/description/) + +```html +[[1,3,1], + [1,5,1], + [4,2,1]] +Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum. +``` + +题目描述：求从矩阵的左上角到右下角的最小路径和，每次只能向左和向下移动。 + +```java +public int minPathSum(int[][] grid) { + if (grid.length == 0 || grid[0].length == 0) return 0; + int m = grid.length, n = grid[0].length; + int[] dp = new int[n]; + for (int i = 0; i < m; i++) { + for (int j = 0; j < n; j++) { + if (j == 0) dp[0] = dp[0] + grid[i][0]; // 只能从上侧走到该位置 + else if (i == 0) dp[j] = dp[j - 1] + grid[0][j]; // 只能从右侧走到该位置 + else dp[j] = Math.min(dp[j - 1], dp[j]) + grid[i][j]; + } + } + return dp[n - 1]; +} +``` + ### 其它问题 **需要冷却期的股票交易**