题目:

给定一个数字三角形,找到从顶部到底部的最小路径和。每一步可以移动到下面一行的相邻数字上。

样例

比如,给出下列数字三角形:

[

     [2],

    [3,4],

   [6,5,7],

  [4,1,8,3]

]

从顶到底部的最小路径和为11 ( 2 + 3 + 5 + 1 = 11)。

注意

如果你只用额外空间复杂度O(n)的条件下完成可以获得加分,其中n是数字三角形的总行数

解题:

求出从顶点到底部所有节点的路径,在选取最小的路径和.这里给的是下三角矩阵,A[i][j] += Max{A[i-1][j-1],A[i-1][j]},对应两个边界的情况:A[i][j] += A[i-1][j]、A[i][j] +=A[i-1][j-1],这样从上向下,在求到最底部时候,找出最小的值。

Java程序:

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.length==1&&triangle[0].length==1)
            return triangle[0][0];
        int minnum = Integer.MAX_VALUE;
      
        for(int i=1;i<triangle.length;i++){
            for(int j=0;j<triangle[i].length;j++){
                if(j==0){
                    triangle[i][j]+=triangle[i-1][j];
                }else if(j==triangle[i].length-1){
                    triangle[i][j]+=triangle[i-1][j-1];
                }else{
                    triangle[i][j]+=Math.min(triangle[i-1][j],triangle[i-1][j-1]);
                }
                if(i==triangle.length-1)
                    minnum = Math.min(minnum,triangle[i][j]);
            }
        }
        return minnum;
    }
}
View Code
总耗时: 2318 ms

Python程序:

class Solution:
    """
    @param triangle: a list of lists of integers.
    @return: An integer, minimum path sum.
    """
    def minimumTotal(self, triangle):
        # write your code here
        
        if triangle==None:
            return 0
        if len(triangle)==1 and len(triangle[0])==1:
            return triangle[0][0]
        minnum = 0 
        m = len(triangle)
        for i in range(1,m):
            for j in range(0,len(triangle[i])):
                if j==0:
                    triangle[i][j] += triangle[i-1][j]
                elif j==len(triangle[i])-1:
                    triangle[i][j] += triangle[i-1][j-1]
                else:
                    triangle[i][j] += min(triangle[i-1][j-1],triangle[i-1][j])
                if i==len(triangle) - 1 and j==0:
                    minnum = triangle[i][j]
                else:
                    minnum = min(minnum,triangle[i][j])
        return minnum
View Code

总耗时: 414 ms

上面是自顶向下的,能否可以自底向上进行,,竟然也可以,这里还不要考虑两个边界的情况,当然上面的其实也可以不考虑的,可以认为是0,这里:triangle[i][j] +=min(triangle[i+1][j] , triangle[i+1][j+1])

Java程序:

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.length==1&&triangle[0].length==1)
            return triangle[0][0];
        int minnum = Integer.MAX_VALUE;
        for(int i=triangle.length-2;i>=0;i--){
            for(int j=0;j<triangle[i].length;j++){
                triangle[i][j] += Math.min(triangle[i+1][j],triangle[i+1][j+1]);
            }
        }
        minnum = triangle[0][0];
        return minnum;
    }
}
View Code

总耗时: 1384 ms

运行时间也少了好多的

Python程序:

class Solution:
    """
    @param triangle: a list of lists of integers.
    @return: An integer, minimum path sum.
    """
    def minimumTotal(self, triangle):
        # write your code here
        
        if triangle==None:
            return 0
        if len(triangle)==1 and len(triangle[0])==1:
            return triangle[0][0]
        minnum = 0 
        m = len(triangle)
        for i in range(m-2,-1,-1):
            for j in range(0,len(triangle[i])):
                triangle[i][j] +=min(triangle[i+1][j] , triangle[i+1][j+1])
        minnum = triangle[0][0]
        return minnum
View Code

总耗时: 439 ms