120. Triangle

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