leetcode 120 Triangle

题目描述

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.

解题思路

因为只有两个方向：下、右下；所以递推公式局部是

dp[m + 1][n] = min(dp[m][n], dp[m][n - 1]) + triangle[m + 1][n]

解题的时候要从一般到特殊，既边缘 case 放在后面考虑（边缘 case 的解决属于思维收敛）。
完整的递推公式如下

if n > 1
    dp[m + 1][n] = min(dp[m][n], dp[m][n - 1]) + triangle[m + 1][n] 
else
    dp[m + 1][0] = dp[m][0] + triangle[m + 1][0]

横向遍历三角形的时候，需要从右向左倒叙地遍历元素，因为 dp table 压缩存储方式成一维空间以后，会面临 dp table 的值被覆盖的问题。

代码实现

class Solution {
public:
    int minimumTotal(const vector<vector<int>>& triangle) const {
    	vector<int> v(triangle.size(), INT_MAX);
    	v[0] = triangle[0][0];
    
    	for (size_t i = 1; i < triangle.size(); i++) {
    		for (int j = i; j >= 0; j--) {
    			switch (j)
    			{
    			case 0:
    				v[0] += triangle[i][0];
    				break;
    			default:
    				v[j] = min(v[j], v[j - 1]) + triangle[i][j];
    				break;
    			}
    		}
    	}
    
    	return *min_element(v.begin(), v.end());
    }
};