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).
代码如下:
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
vector<int> len(triangle.back());
int s = triangle.size();
for (int i = s - 2; i >= 0; i--) {
for (int j = 0; j <= i; j++) {
len[j] = triangle[i][j] + min(len[j], len[j + 1]);
}
}
return len[0];
}
};
解题思路:
- 经分析可得,要求从顶部到底部的最短路径,也即从底部到顶部的最短路径,此问题可以分解为多个子问题:每一层中某个点的值加上一层到顶部的最小值;
- 由于只能走改点相邻的点则,只需比较该点(triangle[i][j])上一层 len[j] 和 len[j + 1] 的大小即可,并遍历第 i 行;
- 重复这个过程即可。