实现代码: ## LeetCode 118classSolution:defgenerate(self,num_rows):## The number of rowstriangle=[]forrow_numinrange(num_rows):## For a specific rowrow=[Nonefor_inrange(row_num+1)]## All None for this rowrow[0]=1## The most left number = 1row[-1]=1## The most right number...
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:...
classSolution{public:vector<int>getRow(introwIndex){ vector<int> ans;for(inti =0; i < rowIndex +1; i++) { ans.push_back(1);for(intj = i -1; j >0; j--) ans[j] += ans[j -1]; }returnans; } };
GivennumRows, generate the firstnumRowsof Pascal's triangle. For example, givennumRows= 5, Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 这是属于基础题目了,记得好像很多基础编程书上都有。 没记错的话,中文名字应该是“杨辉三角”。因为中国人记录这个要早。 本算法...
for x in range(i - 1): row[x+1] = ans[i-1][x] + ans[i-1][x+1] print row ans.append(row) return ans 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. Pascal’s Triangle II
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...
Leetcode: Pascal's Triangle 题目: GivennumRows, generate the firstnumRowsof Pascal's triangle. For example, givennumRows= 5, Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 1. 2. 3. 4. 5. 6. 7.
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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...
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=newArrayList<Integer>();tri.add(1);fo...