Leetcode Triangle

題目描述

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

解題思路

1. 從題目中我們可以知道A[i - 1][n] 和A[i][n] , A[i][n+1]相鄰
2. 因此從題目概述中可以寫出狀態轉移方程
   dp(i, j)表示在第i層（從上往下）的第j個元素中的最短路徑和
   
   dp(i,j)=mindp[i−1][j−1]+A[i][j],dp[i−1][j]+A[i][j](1<=j<=i−1)
   
   
   dp(i,j)=dp[i−1][j]+A[i][j](j=0),dp(i,j)=dp[i−1][j]+A[i][j−1](j=i−1)
3. 答案爲min{dp[n-1][j]}, 0<=j<=i

public:
    int minimumTotal(vector<vector<int>>& triangle) {
        if (triangle.empty() || triangle[0].empty())   return 0;
        int n = triangle.size();
        int dp[n][n];
        int M = 0x01000000;
        memset(dp, 0, sizeof(dp));
        dp[0][0] = triangle[0][0];
        for (int i = 1; i < n; i++) {
            for (int j = 0; j <= i; j++) {
                if (j == 0) {
                    dp[i][j] = dp[i - 1][j] + triangle[i][j];
                }
                else if (j == i) {
                    dp[i][j] = dp[i - 1][j - 1] + triangle[i][j];
                }
                else {
                    dp[i][j] = min(dp[i - 1][j - 1] + triangle[i][j], dp[i - 1][j] + triangle[i][j]);
                }
            }
        }
        for (int i = 0; i < n; i++) {
            M = M > dp[n - 1][i]? dp[n - 1][i]: M;
        }
        return M;
    }
};