120. Triangle

Problem

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.

Solution

数字三角形问题

思路不难，从下往上，递推公式：

dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + a[i][j];

注意的是：

每行元素个数 ＝ 该行是第几行！！

直接滚动数组做了

class Solution { public: int minimumTotal(vector<vector<int>>& triangle) { //不用考虑空的情况 int n = triangle.size(); if (n == 1) { return triangle[0][0]; } vector<int> dp(n); for (int i = 0; i < n; ++i) { dp[i] = triangle[n-1][i]; } for (int i = n-2; i >= 0; --i) { for (int j = 0; j <= i; ++j) { dp[j] = min(dp[j], dp[j+1]) + triangle[i][j]; } } return dp[0]; } };