LeetCode:Triangle
来源:互联网 发布:javascript字符串反转 编辑:程序博客网 时间:2024/04/29 19:36
Given a triangle, find the minimum path sum from top to bottom. Each step you may
move to adjacent numbers on the row below.
For example, given the following triangle
[ [2], [3,4], [6,5,7], [4,1,8,3]]
The minimum path sum from top to bottom is 11
(i.e., 2 + 3 +5 +1 = 11).
Note:
Bonus point if you are able to do this using onlyO(n) extra space, wheren is the total number of
rows in the triangle.
解题思路:
首先dp是很容易想到的,用dp[i][j]表示到第i层j列的最小值,那么状态转移为:
dp[i][j]=min(dp[i-1][j],dp[i-1][j-1])+triangle[i][j] ( 0 < j < triangle[i].size() - 1).由于状态转移只与i和i-1
行有关,所以用滚动数组能将其内存优化为O(n)(此处有常系数2).关于内存的优化,其实是可以做
到严格的O(n)的,我们通过反向更新即可实现.
解题代码:
class Solution {public: int minimumTotal(vector<vector<int> > &triangle) { const int n = triangle.size(); int dp[n]; dp[0] = triangle[0][0]; for (int i = 1; i < n; ++i) { for (int j = triangle[i].size() - 1; j >= 0 ; --j) { if (j == triangle[i].size() - 1 || !j) dp[j] = (!j ? dp[0] : dp[j-1]) + triangle[i][j]; else dp[j] = min(dp[j],dp[j-1]) + triangle[i][j]; } } return *min_element(dp,dp+n); }};
更简单的代码:
class Solution {public: int minimumTotal(vector<vector<int> > &triangle) { const int n = triangle.size(); int dp[n]; for (int i = n -1; i >= 0; --i) dp[i] = triangle[n-1][i]; for (int i = n -2; i >= 0; --i) for (int j = 0; j < triangle[i].size(); ++j) dp[j] = min(dp[j],dp[j+1]) + triangle[i][j]; return dp[0]; }};
0 0
- LeetCode : Triangle
- [LeetCode] Triangle
- [Leetcode] Triangle
- [LeetCode] Triangle
- 【leetcode】Triangle
- LeetCode - Triangle
- [LeetCode]Triangle
- [Leetcode]Triangle
- [leetcode]Triangle
- Leetcode: Triangle
- [LeetCode] Triangle
- LeetCode-Triangle
- [leetcode] Triangle
- [LeetCode] Triangle
- LeetCode - Triangle
- 【Leetcode】Triangle
- LeetCode | Triangle
- [leetcode]Triangle
- Oracle中时间处理_tochar
- java学习笔记(2)
- python 遇到 syntaxerror: non-ascii character问题
- 选课时间(题目已修改,注意读题)
- Android 开源框架Universal-Image-Loader完全解析(一)--- 基本介绍及使用
- LeetCode:Triangle
- KM算法 HDOJ1533 匈牙利算法
- Android 开源框架Universal-Image-Loader完全解析(二)--- 图片缓存策略详解
- 黑马程序员——Java基础---网络编程
- mjpg-streamer项目源码分析 2
- Android-03-模拟器
- D_DB2 DPF环境搭建
- 转:TestNG官方文档中文版(02)—— Annotation
- xtrabackup 备份mysql数据库三: innobackupex 测试一个全量和两个增量的备份恢复测试