Triangle

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

  1. [
  2.      [2],
  3.     [3,4],
  4.    [6,5,7],
  5.   [4,1,8,3]
  6. ]

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.

分析

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)f(i,j)=\min\left{f(i+1,j),f(i+1,j+1)\right}+(i,j)

代码

  1. // Triangle
  2. // 时间复杂度O(n^2)，空间复杂度O(1)
  3. class Solution {
  4. public:
  5.     int minimumTotal (vector<vector<int>>& triangle) {
  6.         for (int i = triangle.size() - 2; i >= 0; --i)
  7.             for (int j = 0; j < i + 1; ++j)
  8.                 triangle[i][j] += min(triangle[i + 1][j],
  9.                         triangle[i + 1][j + 1]);
  11.         return triangle [0][0];
  12.     }
  13. };

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/dp/triangle.html