[] Data structures and algorithms dynamic programming - and minimum path (common matrix, triangle two questions)

And minimum path

LeetCode: minimum path and

Subject description:

Given a non-negative integer of mxn grid, find a path from left to bottom right, so that the sum of the minimum number of paths.

Note: you can only move one step down or to the right.

Example:

输入:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
输出: 7
解释: 因为路径 1→3→1→1→1 的总和最小。

thought:

Dynamic programming, you can use the original array as an array dp

Code:

class Solution {
    public int minPathSum(int[][] grid) {
        int i=0,j=0;
        for(i=0;i<grid.length;++i){
            for(j=0;j<grid[0].length;++j){
                if(i>0&&j>0){
                    grid[i][j]+= Math.min(grid[i-1][j],grid[i][j-1]);
                }else{
                    grid[i][j]+= (i==0?0:grid[i-1][j]) + (j==0?0:grid[i][j-1]);
                }
            }
        }
        return grid[i-1][j-1];
    }
}

Triangles and minimum path

LeetCode: triangles, and minimum path

Subject description:

Given a triangle, find the minimum and the top-down path. Each step can move to the next line adjacent nodes.

If you can use only O (n) extra space (n number of rows as a triangle) to solve this problem, then your algorithm will be a plus.

Example:

例如给定三角形
[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]
自顶向下的最小路径和为 11（即，2 + 3 + 5 + 1 = 11）。

thought:

From the bottom up, modify array dp

Code:

The first method: modified on the source array. Such seemingly inefficient.

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        for(int i=triangle.size()-2;i>=0;--i){
            for(int j=0;j<i+1;++j){
                triangle.get(i).set(j,triangle.get(i).get(j)+Math.min(triangle.get(i+1).get(j+1),triangle.get(i+1).get(j)));
            }
        }
        return triangle.get(0).get(0);
    }
}

The second method: set up an array of dp, dp modify the array; note that this is very clever, every modification will not affect the judgment of the next cycle; second, each cycle, the last number will not change it until the last round , plus the number of the top, the final result dp [0].

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int len = triangle.size();
        int[] dp=new int[len];
        for(int i=0;i<len;++i){
            dp[i]=triangle.get(len-1).get(i);
        }
        for(int i=len-2;i>=0;--i){
            for(int j=0;j<i+1;++j){
                dp[j] = Math.min(dp[j],dp[j+1]) + triangle.get(i).get(j);
            }
        }
        return dp[0];
    }
}