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];
}
};