Triangle -- C3

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

 Notice

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.

Have you met this question in a real interview? Yes
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).

Tags 
Dynamic Programming
Related Problems 
Easy Minimum Path Sum

矩阵的题常构造新的矩阵转化为题意, 简单的记忆化存储, 常用dfs的分治策略自底向下回溯, 找一条最小的最合适的路径

递归出口(出范围, 遍历过, 满足题意(到节点) ) , 递归条件(对邻居的判断, 遍历过, 出范围, 邻居), 返回值常用分治法(后序) 尾递归,递的时候是用的上面的值, 回的时候是用的下面的值.

自底向上

记忆化搜索: 分治法加入了数组变量
 　　

public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null || triangle.size() == 0) {
            return 0;
        }
      
        long[][] dp = new long[triangle.size()][triangle.size()];
       for (long[] d : dp) {
        Arrays.fill(d, Long.MAX_VALUE);
           
       }
       
        for (int i = 0 ; i < dp[0].length; i++) {
            dp[dp.length - 1][i] = triangle.get(triangle.size() - 1).get(i);
        }
        dfs(triangle, dp, 0, 0);
        return (int)dp[0][0];

    }
    private int dfs(List<List<Integer>> triangle, long[][] dp, int i, int j) {
        if (dp[i][j] != Long.MAX_VALUE) {
            return (int)dp[i][j];
        }
        
        int ans = triangle.get(i).get(j);

        int next = Math.min(dfs(triangle, dp, i + 1, j),dfs(triangle, dp, i+1, j + 1)) + ans;
        dp[i][j] = next;
        return next;
    }

　　

动归, 动归矩阵, 自顶向下

public class Solution {
    /**
     * @param triangle: a list of lists of integers.
     * @return: An integer, minimum path sum.
     */
    public int minimumTotal(int[][] triangle) {
        // write your code here
        if (triangle == null || triangle.length == 0) {
            return -1;
        }
        
        // state:
        int n = triangle.length;
        int[][] f = new int [n][n];
        
        //initialize
        f[0][0] = triangle[0][0];
        for (int i = 1; i < n; i++) {
            f[i][0] = f[i - 1][0] + triangle[i][0];
            f[i][i] = f[i - 1][i - 1] + triangle[i][i];
        }
        
        // top down
        
        for (int i = 1; i < n; i++) {
            for (int j = 1; j < i; j++) {
                f[i][j] = Math.min(f[i - 1][j], f[i - 1][j - 1]) + triangle[i][j];
                
            }
        }
        
        //answer
        int ans = f[n - 1][0];
        for (int i = 1; i < n; i++) {
            if (ans > f[n - 1][i]) {
                ans = f[n - 1][i];
            }
        }
        
        return ans;
}