LeetCode Triangle

来源：互联网发布：sql中replace函数用法编辑：程序博客网时间：2024/05/23 18:33

题目得说明路径不能跨越，否则的话，每层找最小的就可以了。为什么多提交两次呢？dp[j] = (dp[j]>dp[j+1]?dp[j+1]:dp[j]) + triangle[i][j];要是条件表达式不加括号的话不可以，那样的话后面的加法就没加上。还有一点就是j从小到大遍历，否则的话会覆盖结果，正常人都会从小到大。

// LeetCode_Triangle.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>#include <vector>using namespace std;int minimumTotal(vector<vector<int> > &triangle) {int row = triangle.size();int column;int ret = 0;if(row!=0)column = triangle[row-1].size();elsereturn ret;int *dp = new int[column];for(int i=0;i<column;i++)dp[i] = triangle[row-1][i];for (int i=row-2;i>=0;i--){for (int j=0;j<=i;j++){dp[j] = (dp[j]>dp[j+1]?dp[j+1]:dp[j]) + triangle[i][j];}}ret = dp[0];delete[] dp;return ret;}int _tmain(int argc, _TCHAR* argv[]){vector<int> temp;vector<vector<int> > tri;temp.push_back(-1);tri.push_back(temp);temp.clear();temp.push_back(2);temp.push_back(3);tri.push_back(temp);temp.clear();temp.push_back(1);temp.push_back(-1);temp.push_back(-1);tri.push_back(temp);cout<<minimumTotal(tri);system("pause");return 0;}

0 0