Github 同步地址: https://github.com/grandyang/leetcode/issues/813 参考资料: https://leetcode.com/problems/largest-sum-of-averages/ https://leetcode.com/problems/largest-sum-of-averages/solution/ https://leetcode.com/problems/largest-sum-of-averages/discuss/122739/C++JavaPython-Easy-Understood...
原题链接在这里:https://leetcode.com/problems/largest-sum-of-averages/ 题目: We partition a row of numbersAinto at mostKadjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use eve...
813. Largest Sum of Averages We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not...
LeetCode 813.LargestSumof Averages 原题链接在这里:https://leetcode.com/problems/largest-sum-of-averages/ 题目: We partition a row of numbers A into at most K adjacent (non-empty) groups, then ou LeetCode DP i++ java JAVA 转载 ...
Leetcode Solutions (0x6A73). Contribute to waffle87/leetcode development by creating an account on GitHub.
code 代码语言:javascript 代码运行次数:0 运行 AI代码解释 type TreeNode struct { Val int Left *TreeNode Right *TreeNode } var res []int func largestValues(root *TreeNode) []int { res = []int{} preorder(root, 0) return res } func preorder(root *TreeNode, d int) { if root == ...
package leetcode func largestPerimeter(A []int) int { if len(A) < 3 { return 0 } quickSort164(A, 0, len(A)-1) for i := len(A) - 1; i >= 2; i-- { if (A[i]+A[i-1] > A[i-2]) && (A[i]+A[i-2] > A[i-1]) && (A[i-2]+A[i-1] > A[i]) { retu...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算
2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. Refernce: http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-813-largest-sum-of-averages/ 永远渴望,大智若愚(stay hungry, stay foolish)
这道题让我想起了算法导论上的一道切绳子?的题。可以用二维dp去做,dp[i][j]表示将nums中前i个元素split成j段的maxSum。 注意dp[][]的初始化! classSolution{publicintsplitArray(int[]nums,intm){intn=nums.length;int[][]dp=newint[n+1][m+1];for(int[]row:dp){Arrays.fill(row,Integer.MAX_...