Leetcode-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.
求数字三角形从顶部到底部和最小的一条路径，数字移动只能往左下或右下。

思路

动态规划思想，设状态为f(i, j)，表示从从位置(i,j)出发，路径的最小和，则状态转移方程为

f(i, j) = min{f(i+1, j), f(i+1, j+1)} + (i, j);

代码

方法1：不修改原数组，需要O(n)额外存储空间

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        int dp[2][triangle.size()];
        int flag = 0;

        for(int i=0; i<triangle.size(); i++)
            dp[flag][i] = triangle[triangle.size()-1][i];

        for(int i=triangle.size()-2; i>=0; i--) {
            int t = (flag+1)%2;
            for(int j=0; j<=i; j++)
                dp[t][j] = min(dp[flag][j], dp[flag][j+1]) + triangle[i][j];
            flag = t;
        }
        return dp[flag][0];
    }
};

方法2：不需要额外存储空间，需要修改原数组

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        for (int i = triangle.size() - 2; i >= 0; --i)
            for (int j = 0; j < i + 1; ++j)
                triangle[i][j] += min(triangle[i+1][j], triangle[i+1][j+1]);
        return triangle[0][0];
    }
};