检查矩阵内部的总和，并以递归方式保留第二个数组上的路径

Question

这个问题是我在Java学期末的测试中提出的：

给定一个正数(未排序(矩阵m，一个整数sum和另一个矩阵p，到处都是0。 递归检查m内部是否有路径，其总和将等于sum。

规则：

您只能在数组中向下、向上、向左或向右移动。

找到路径后，矩阵p将填充正确路径上的1's。

只有 1 条路径

该方法完成后，应0p上的所有其他单元格。

如果没有通往这笔钱的途径，你将离开p因为你得到了他。

例：

int[][] p = {{0,0,0,0},
{0,0,0,0},
{0,0,0,0},
{0,0,0,0}};

一开始。

矩阵为：

int [][] hill = {{3,8,7,1},
{5,15,2,4},
{12,14,-13,22},
{13,16,17,52}};

如果sum = 23调用该方法，该方法将返回 true，p为：

int[][] p = {{1,0,0,0},
{1,1,0,0},
{0,0,0,0},
{0,0,0,0}};

该方法必须是递归的

这个问题简直让测试变得像地狱一样......

希望你能弄清楚，也许可以帮助我理解它！！谢谢

我的进度：

public static boolean findSum(int[][] mat , int sum , int[][]path){
return findSum(mat,sum,path,0,0);
}
private static boolean findSum(int[][] m, int sum, int[][] p, int i, int j) {
if (i>=m.length || j>= m[i].length) return false;

boolean op1 = finder(m,sum-m[i][j],p,i,j);
boolean op2 = findSum(m,sum,p,i+1,j);
boolean op3 = findSum(m,sum,p,i,j+1);
if (op1) return true;
else if (op2) return true;
return op3;
}
private static boolean finder(int[][] m, int sum,int[][]p , int i, int j) {
if (sum==0) {
p[i][j]=1;
return true;
}
p[i][j]=1;
boolean op1=false,op2=false,op3=false,op4=false;
if (i>0 && p[i-1][j]==0 && sum-m[i][j]>=0) op1 = finder(m, sum - m[i][j], p, i - 1, j);
if (i<m.length-1 && p[i+1][j]==0&& sum-m[i][j]>=0) op2 = finder(m, sum - m[i][j], p, i + 1, j);
if (j>0 && p[i][j-1]==0&& sum-m[i][j]>=0) op3 = finder(m, sum - m[i][j], p, i, j - 1);
if (j<m[i].length-1 && p[i][j+1]==0&& sum-m[i][j]>=0) op4 = finder(m, sum - m[i][j], p, i, j + 1);
else p[i][j]=0;
return op1||op2||op3||op4;
}

Answer 1

我真的很喜欢解决这个问题。我已经用python完成了它，但你可以很容易地将其扩展到Java。我已经注释了代码供您理解。如果有什么你没有得到的或可以改进的，请告诉我。

顺便说一句，在您的示例中，一个总和有多个路径，下面的代码可以找到所有路径。

hill = [[3,8,7,1],[5,15,2,4],[12,14,-13,22],[13,16,17,52]]
p = [ [0 for x in range (4)] for y in range(4)]
num = 23
def checkPath(p, r, c): #Check boundaries
res = []
if r+1<len(p):
res.append(p[r+1][c] == 0)
if r-1>=0:
res.append(p[r-1][c] == 0)
if c+1<len(p[0]):
res.append(p[r][c+1] == 0)
if c-1>=0:
res.append(p[r][c-1] == 0)
return res

def pathF(tot, hill, p, r, c):
p[r][c] = 1 #mark visited
tot = tot + hill[r][c]    #update total
if tot == num: #solution found
print("Found", p)
else:
if any (checkPath(p,r,c)):
if r+1<len(p) and p[r+1][c] == 0: #move right checking if it wasnt visited
pathF(tot,hill,p,r+1,c)
if r-1>=0 and p[r-1][c] == 0:
pathF(tot,hill,p,r-1,c)
if c+1<len(p[0]) and p[r][c+1] == 0:
pathF(tot,hill,p,r,c+1)
if c-1>=0 and p[r][c-1] == 0:
pathF(tot,hill,p,r,c-1)
p[r][c]=0 #mark unvisited
tot = tot - hill[r][c]     #set total to original       

for x in range(len(hill)): #iterate over every starting point possible
for y in range(len(hill[0])):
pathF(0,hill,p,x,y)

这是 num = 23 的输出

Found [[1, 0, 0, 0], [1, 0, 1, 0], [1, 1, 1, 0], [0, 0, 0, 0]]
Found [[1, 0, 0, 0], [1, 1, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Found [[1, 1, 1, 1], [0, 0, 0, 1], [0, 0, 0, 0], [0, 0, 0, 0]]
Found [[0, 1, 0, 0], [0, 1, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Found [[0, 1, 1, 1], [0, 0, 1, 1], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[1, 1, 1, 0], [1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 0, 1], [0, 1, 1, 1], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 0, 1], [0, 1, 1, 1], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 0], [1, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[1, 1, 1, 0], [1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Found [[0, 0, 0, 0], [1, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 0, 0], [1, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 0, 1], [0, 1, 1, 1], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 1, 0, 0], [0, 1, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Found [[1, 0, 0, 0], [1, 1, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 0, 0], [1, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[1, 0, 0, 0], [1, 0, 1, 0], [1, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[1, 1, 1, 1], [0, 0, 0, 1], [0, 0, 0, 0], [0, 0, 0, 0]]
Found [[0, 0, 0, 0], [0, 0, 1, 1], [0, 1, 1, 0], [0, 1, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 0], [1, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 1, 1, 1], [0, 0, 1, 1], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 0, 0], [1, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 0, 0], [0, 0, 0, 0], [0, 1, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 0, 1], [0, 1, 1, 1], [0, 1, 1, 0], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 0, 0], [0, 0, 0, 0], [0, 1, 1, 1], [0, 0, 0, 0]]
Found [[0, 0, 0, 0], [0, 0, 1, 1], [0, 1, 1, 0], [0, 1, 0, 0]]

Answer 2

所以我设法让它工作:) 在我的课程讲师的帮助下，这是一个完整的 Java 解决方案！

public static boolean findSum(int[][] m ,int s, int[][]p){
return findSum(m,s,p,0,0); //calling overloading
}
private static boolean findSum(int[][] m, int s, int[][] p, int i, int j) {
if (i<0 || i>=m.length) return false; //stop condition
if (finder(m,s,p,i,j)) return true; //first recursion
if (j<m[0].length-1) //if the iterator is not on the end of the row 
return  findSum(m,s,p,i,j+1); //recursive call 
else //if i checked the whole row , the column will be increase.
return findSum(m,s,p,i+1,0);//recursive call
}
private static boolean finder(int[][] m, int s, int[][] p, int i, int j) {
if (s == 0) return true;
if (i<0 || i>=m.length || j<0 || j>=m.length || s<0 || p[i][j] == 1) return false;
p[i][j] =1;
boolean u=false,d=false,r=false,l=false;
if (i>0) u = finder(m,s-m[i][j],p,i-1,j);
if (i<m.length-1) d = finder(m,s-m[i][j],p,i+1,j);
if (j>0) l = finder(m,s-m[i][j],p,i,j-1);
if (i<m.length-1) r = finder(m,s-m[i][j],p,i,j+1);
if (u) return true;
else if (d) return true;
else if (r) return true;
else if (l) return true;
else {
p[i][j] = 0;
return false;
}
}