From ebb3ce62d3e681557f1b84492641dff19ff56dad Mon Sep 17 00:00:00 2001 From: CyC2018 <1029579233@qq.com> Date: Thu, 14 Jun 2018 10:18:14 +0800 Subject: [PATCH] auto commit --- "notes/Leetcode \351\242\230\350\247\243.md" | 85 +++++++++++++------- 1 file changed, 57 insertions(+), 28 deletions(-) diff --git "a/notes/Leetcode \351\242\230\350\247\243.md" "b/notes/Leetcode \351\242\230\350\247\243.md" index f5e01cb2..8c1fd1f2 100644 --- "a/notes/Leetcode \351\242\230\350\247\243.md" +++ "b/notes/Leetcode \351\242\230\350\247\243.md" @@ -3116,7 +3116,9 @@ the contiguous subarray [4,-1,2,1] has the largest sum = 6. ```java public int maxSubArray(int[] nums) { - if (nums == null || nums.length == 0) return 0; + if (nums == null || nums.length == 0) { + return 0; + } int preSum = nums[0]; int maxSum = preSum; for (int i = 1; i < nums.length; i++) { @@ -3142,7 +3144,9 @@ dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。 ```java public int numberOfArithmeticSlices(int[] A) { - if (A == null || A.length == 0) return 0; + if (A == null || A.length == 0) { + return 0; + } int n = A.length; int[] dp = new int[n]; for (int i = 2; i < n; i++) { @@ -3150,9 +3154,11 @@ public int numberOfArithmeticSlices(int[] A) { dp[i] = dp[i - 1] + 1; } } - int ret = 0; - for (int cnt : dp) ret += cnt; - return ret; + int total = 0; + for (int cnt : dp) { + total += cnt; + } + return total; } ``` @@ -3176,16 +3182,21 @@ public int minDistance(String word1, String word2) { int[][] dp = new int[m + 1][n + 1]; for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) { - if (i == 0 || j == 0) continue; - dp[i][j] = word1.charAt(i - 1) == word2.charAt(j - 1) ? - dp[i - 1][j - 1] + 1 : Math.max(dp[i][j - 1], dp[i - 1][j]); + if (i == 0 || j == 0) { + continue; + } + if (word1.charAt(i - 1) == word2.charAt(j - 1)) { + dp[i][j] = dp[i - 1][j - 1] + 1; + } else { + dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]); + } } } return m + n - 2 * dp[m][n]; } ``` -**修改一个字符串成为另一个字符串** +**编辑距离** [72. Edit Distance (Hard)](https://leetcode.com/problems/edit-distance/description/) @@ -3194,7 +3205,7 @@ Example 1: Input: word1 = "horse", word2 = "ros" Output: 3 -Explanation: +Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e') @@ -3202,7 +3213,7 @@ Example 2: Input: word1 = "intention", word2 = "execution" Output: 5 -Explanation: +Explanation: intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') @@ -3210,6 +3221,8 @@ exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u') ``` +题目描述：修改一个字符串成为另一个字符串，使得修改次数最少。一次修改操作包括：插入一个字符、删除一个字符、替换一个字符。 + ```java public int minDistance(String word1, String word2) { if (word1 == null || word2 == null) { @@ -3248,8 +3261,8 @@ public int minDistance(String word1, String word2) { 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++) { + 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))); } } @@ -3270,7 +3283,9 @@ public int numSquares(int n) { for (int i = 1; i <= n; i++) { int min = Integer.MAX_VALUE; for (int square : squareList) { - if (square > i) break; + if (square > i) { + break; + } min = Math.min(min, dp[i - square] + 1); } dp[i] = min; @@ -3299,17 +3314,25 @@ private List generateSquareList(int n) { ```java public int numDecodings(String s) { - if(s == null || s.length() == 0) return 0; + 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++) { + 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; + 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]; + if (two <= 26) { + dp[i] += dp[i - 2]; + } } return dp[n]; } @@ -3342,8 +3365,8 @@ public int uniquePaths(int m, int n) { ```java public int uniquePaths(int m, int n) { - int S = m + n - 2; // 总共的移动次数 - int D = m - 1; // 向下的移动次数 + int S = m + n - 2; // 总共的移动次数 + int D = m - 1; // 向下的移动次数 long ret = 1; for (int i = 1; i <= D; i++) { ret = ret * (S - D + i) / i; @@ -3367,14 +3390,20 @@ Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes ```java public int minPathSum(int[][] grid) { - if (grid.length == 0 || grid[0].length == 0) return 0; + 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]; + 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]; @@ -3517,11 +3546,11 @@ public int minSteps(int n) { ### 素数 -（一）素数分解 +**素数分解** 每一个数都可以分解成素数的乘积，例如 84 = 22 \* 31 \* 50 \* 71 \* 110 \* 130 \* 170 \* … -（二）整除 +**整除** 令 x = 2m0 \* 3m1 \* 5m2 \* 7m3 \* 11m4 \* … @@ -3529,7 +3558,7 @@ public int minSteps(int n) { 如果 x 整除 y（y mod x == 0），则对于所有 i，mi <= ni。 -（三）最大公约数最小公倍数 +**最大公约数最小公倍数** x 和 y 的最大公约数为：gcd(x,y) = 2min(m0,n0) \* 3min(m1,n1) \* 5min(m2,n2) \* ... -- GitLab