这个问题是我期末 Java 测试中的问题:
给定正数矩阵(未排序)m
,一个整数 sum
另一个矩阵p
充满0
遍。
递归检查 m
内是否有路径它的总和将等于 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1's
走在正确的道路上。
只有1条路径
p
上的所有其他单元格应该是0
该方法完成后。
如果没有路径到达此金额,您将离开 p
当你得到他时。
示例:
int[][] p = {{0,0,0,0},
{0,0,0,0},
{0,0,0,0},
{0,0,0,0}};
一开始。
矩阵是:
int [][] hill = {{3,8,7,1},
{5,15,2,4},
{12,14,-13,22},
{13,16,17,52}};
如果您在 sum = 23
上调用该方法该方法将返回 true 和 p
将是:
int[][] p = {{1,0,0,0},
{1,1,0,0},
{0,0,0,0},
{0,0,0,0}};
方法必须是递归的
这个问题让考试变得很糟糕......
希望你能弄清楚并帮助我理解它!谢谢
我的进展:
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;
}
最佳答案
我真的很喜欢解决这个问题。我已经用 python 完成了它,但你可以轻松地将其扩展到 Java。我已对代码进行了注释以便您理解。如果有什么您没有得到或可以改进的地方,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]]
关于java - 检查矩阵内的总和并递归地将路径保留在第二个数组上,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51032518/