Java for LeetCode 119 Pascal's Triangle II Given an indexk, return thekthrow of the Pascal's triangle. For example, givenk= 3, Return[1,3,3,1]. 解题思路: 注意,本题的k相当于上题的k+1,其他照搬即可,JAVA实现如下: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 publicList<In...
Given an integer numRows, return the first numRows of Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it as shown: Example 1: Input: numRows = 5 Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]] Example 2: Input:...
Given a non-negative integer numRows, generate the first numRows of Pascal's triangle. 1. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input: 5 Output: [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 1. 2. 3. 4...
Given numRows, generate the first numRows of Pascal’s triangle. For example, given numRows = 5, Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 这个就是中国最伟大的杨辉三角形的问题。 代码如下: import java.util.ArrayList; import java.util.List; /* * 中国最...
杨辉三角还是 Pascal Triangle 杨辉三角在英文版的LeetCode中被称为“Pascal Triangle”,这不免让人疑惑,这都是是谁的三角呢。 要不这样,谁生的早,就算谁的。 杨辉,生于1238年: Pascal(Blaise Pascal?),生于1623: OK,杨辉胜出 -- 那就叫杨辉三角。
sort() # dp[i] 表示第 i 个孩子分到的糖果数,初始化最少每人一个 dp: List[int] = [1] * n # ans 维护所有孩子分到的糖果数之和 ans: int = 0 #按 rating 升序进行状态转移,这样就能保证在更新 dp[i] 时, # 其左右两侧的 dp 值均已确定 for (rating, i) in cells: # 如果其评分...
119 119. Pascal's Triangle II.java Easy [Array, Basic Implementation] O(k^2), pascal triangle size O(k^2) Java 412 1197 1197. Minimum Knight Moves.java Medium [BFS] O(8^n) O(8^n) Java 413 493 493. Reverse Pairs.java Medium [BST, Binary Indexed Tree, Divide and Conquer, Merge...
Anyway, Good luck, Richardo! My code: publicclassSolution{publicList<List<Integer>>generate(intnumRows){ArrayList<List<Integer>>ret=newArrayList<List<Integer>>();if(numRows<=0)returnret;ArrayList<Integer>tri=newArrayList<Integer>();tri.add(1);ret.add(tri);for(inti=1;i<numRows;i++){tri...
0116-Populating-Next-Right-Pointers-in-Each-Node/cpp-0116 0117-Populating-Next-Right-Pointers-in-Each-Node-II/cpp-0117 0117-Populating-Next-Right-Pointers-in-Each-Node-II/cpp-0117 0118-Pascals-Triangle/cpp-0118 0118-Pascals-Triangle/cpp-0118 0119-Pascals-Triangle-II/cpp-0119 0119-Pascals-Triang...
升级版本,注意 Pascal's Triangle, 其实是左右是对称的. 在子函数pascal不用全部都遍历计算,只用处理len/2,其他用对称就可以 vector<int>pascal(vector<int>&nums,intline){vector<int>rows(line+1);intlen=nums.size();rows[0]=1;for(inti=1;i<=len/2;++i){rows[i]=nums[i-1]+nums[i];rows[le...