120. Triangle

题目

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 only O(n) extra space, where n is the total number of rows in the triangle. 

解析

• 注意思路，从上往下，或者从下往上都可以，当遇见有覆盖的现象，考虑逆序处理。

// Triangle
class Solution_120 {
public:
    // top-down 
    int minimumTotal1(vector<vector<int>>& triangle) {
        vector<int> res(triangle.size(), triangle[0][0]);

        for (unsigned int i = 1; i < triangle.size(); i++)
            for (int j = i; j >= 0; j--) {
            if (j == 0)
                res[0] += triangle[i][j];
            else if (j == i)
                res[j] = triangle[i][j] + res[j - 1];
            else
                res[j] = triangle[i][j] + min(res[j - 1], res[j]);
            }
        return *min_element(res.begin(), res.end());
    }

    // bottom-up
    int minimumTotal2(vector<vector<int>>& triangle) {

        vector<int> res = triangle.back();
        for (int i = triangle.size() - 2; i >= 0; i--)
            for (unsigned int j = 0; j <= i; j++)
                res[j] = triangle[i][j] + min(res[j], res[j + 1]);
        return res[0];
    }

    int minimumTotal(vector<vector<int>>& triangle) {
        int row = triangle.size();
        
        if (row==0)
        {
            return 0;
        }
        if (row==1)
        {
            return triangle[0][0];
        }
        int ret = 0;
        

        vector<int> vec(triangle.size(), triangle[0][0]); //初始化

        for (int i = 1; i < row;i++) //当前行
        {
            for (int j = 0; j < triangle[i].size(); j++)
            {
                if (j==0)
                {
                    vec[j] = vec[j] + triangle[i][j];
                }else if (j==triangle[i].size()-1)
                {
                    vec[j] = vec[j-1] + triangle[i][j]; //bug 会叠加上一次改变的值 //变顺序啊！！！逆序
                }
                else
                {
                    vec[j] = triangle[i][j] + min(vec[j - 1], vec[j]);
                }
            }
        }
        return *min_element(vec.begin(), vec.end());
    }
};

• 测试输入

// 修改输入样例:
//    4
 
//    2
//    3 4
//    6 5 7
//    4 1 8 3

int main()
{
    ifstream infile("in.txt", ifstream::in);
    #define cin infile
    vector<vector<int> > triangle;
    int n;
    cin >> n;
    for(int i = 0; i < n; i++) {
        vector<int> vi(i+1, 0);
        for(int j = 0; j < i+1; j++)
            cin >> vi[j];
        triangle.push_back(vi);
    }
    Solution* s = new Solution();
    cout << s->minimumTotal(triangle) << endl;
    return 0;
}


#include <iostream>
#include <cstdio>

using namespace std;

int main(){
    freopen("1.in", "r", stdin);
    int n, ans = 0;
    cin >> n;
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            int x; scanf("%d", &x);
            ans += x;
        }
    }
    cout << ans << endl;
    return 0;
}

题目来源

• 120. Triangle