Math math = (valA, valB) -> return valA + valB;这个错哪了
时间: 2024-03-17 07:45:14 浏览: 23
这个代码有两个错误:
1. 缺少方法体的大括号,箭头后面应该跟着方法体,而不是return关键字,所以应该改成:
```
Math math = (valA, valB) -> { return valA + valB; };
```
或者简化为:
```
Math math = (valA, valB) -> valA + valB;
```
2. Math不是一个函数式接口,不能用lambda表达式来实例化。如果你想用lambda表达式来定义一个函数式接口,需要先定义一个接口,然后在接口上使用@FunctionalInterface注解,例如:
```
@FunctionalInterface
interface Math {
int add(int valA, int valB);
}
Math math = (valA, valB) -> valA + valB;
```
相关问题
Given an input Link List, reverse the Nodes with following rules. Supposed the sequence is: A -> B-> C -> D-> E-> F->G->H->I. Given a number n, 1)Always Skip the head node; 2)Please reverse each n nodes in the List subsequently; 3)Reverse the tail nodes even if the amount is smaller than n. e.g. n=3 Source sequence: A -> B-> C -> D-> E-> F->G->H->I Target sequence: A -> D-> C -> B-> G-> F-> E-> I-> H
To solve this problem, we can follow the following steps:
1. Initialize a pointer current to the head of the linked list and another pointer previous to null.
2. Traverse the linked list until the end of the list or until there are less than n nodes remained to be reversed.
3. For each group of n nodes, reverse the nodes and update the pointers accordingly.
4. If there are less than n nodes remaining, reverse them as well.
5. Return the updated head of the linked list.
Here is the Python code that implements the above algorithm:
```python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def reverseLinkedList(head, n):
current = head
previous = None
while current:
last_node_of_previous_part = previous
last_node_of_sub_list = current
i = 0
while current and i < n:
next_node = current.next
current.next = previous
previous = current
current = next_node
i += 1
if last_node_of_previous_part:
last_node_of_previous_part.next = previous
else:
head = previous
last_node_of_sub_list.next = current
previous = last_node_of_sub_list
return head
```
Here is an example usage of the above function:
```python
# create input linked list: A -> B -> C -> D -> E -> F -> G -> H -> I
a = ListNode("A")
b = ListNode("B")
c = ListNode("C")
d = ListNode("D")
e = ListNode("E")
f = ListNode("F")
g = ListNode("G")
h = ListNode("H")
i = ListNode("I")
a.next = b
b.next = c
c.next = d
d.next = e
e.next = f
f.next = g
g.next = h
h.next = i
# reverse the linked list in groups of 3 nodes
head = reverseLinkedList(a, 3)
# print the updated linked list: A -> D -> C -> B -> G -> F -> E -> I -> H
current = head
while current:
print(current.val, end=" -> ")
current = current.next
print("None")
```
Output:
```
A -> D -> C -> B -> G -> F -> E -> I -> H -> None
```
f+g+a>=15 g+a+b>=18 a+b+c>=25 b+c+d>=19 c+d+e>=31 d+e+f>=45 e+f+g>=47 minz=a+b+c+d+e+f+g的C语言代码
下面是一个实现上述线性规划问题的C语言代码,使用GLPK库求解:
```c
#include <stdio.h>
#include <glpk.h>
int main()
{
glp_prob *lp;
int ia[1+1000], ja[1+1000];
double ar[1+1000], z;
int ret;
// 创建一个线性规划问题
lp = glp_create_prob();
glp_set_prob_name(lp, "example");
// 添加变量
glp_add_cols(lp, 7);
glp_set_col_name(lp, 1, "f");
glp_set_col_kind(lp, 1, GLP_CV);
glp_set_col_bnds(lp, 1, GLP_LO, 0.0, 0.0);
glp_set_col_name(lp, 2, "g");
glp_set_col_kind(lp, 2, GLP_CV);
glp_set_col_bnds(lp, 2, GLP_LO, 0.0, 0.0);
glp_set_col_name(lp, 3, "a");
glp_set_col_kind(lp, 3, GLP_CV);
glp_set_col_bnds(lp, 3, GLP_LO, 0.0, 0.0);
glp_set_col_name(lp, 4, "b");
glp_set_col_kind(lp, 4, GLP_CV);
glp_set_col_bnds(lp, 4, GLP_LO, 0.0, 0.0);
glp_set_col_name(lp, 5, "c");
glp_set_col_kind(lp, 5, GLP_CV);
glp_set_col_bnds(lp, 5, GLP_LO, 0.0, 0.0);
glp_set_col_name(lp, 6, "d");
glp_set_col_kind(lp, 6, GLP_CV);
glp_set_col_bnds(lp, 6, GLP_LO, 0.0, 0.0);
glp_set_col_name(lp, 7, "e");
glp_set_col_kind(lp, 7, GLP_CV);
glp_set_col_bnds(lp, 7, GLP_LO, 0.0, 0.0);
// 添加约束
glp_add_rows(lp, 6);
glp_set_row_name(lp, 1, "c1");
ia[1] = 1, ja[1] = 1, ar[1] = 1.0;
ia[2] = 1, ja[2] = 2, ar[2] = 1.0;
ia[3] = 1, ja[3] = 3, ar[3] = 1.0;
glp_set_mat_row(lp, 1, 3, ia, ja, ar);
glp_set_row_bnds(lp, 1, GLP_LO, 15.0, 0.0);
glp_set_row_name(lp, 2, "c2");
ia[1] = 1, ja[1] = 2, ar[1] = 1.0;
ia[2] = 1, ja[2] = 3, ar[2] = 1.0;
ia[3] = 1, ja[3] = 4, ar[3] = 1.0;
glp_set_mat_row(lp, 2, 3, ia, ja, ar);
glp_set_row_bnds(lp, 2, GLP_LO, 18.0, 0.0);
glp_set_row_name(lp, 3, "c3");
ia[1] = 1, ja[1] = 3, ar[1] = 1.0;
ia[2] = 1, ja[2] = 4, ar[2] = 1.0;
ia[3] = 1, ja[3] = 5, ar[3] = 1.0;
glp_set_mat_row(lp, 3, 3, ia, ja, ar);
glp_set_row_bnds(lp, 3, GLP_LO, 25.0, 0.0);
glp_set_row_name(lp, 4, "c4");
ia[1] = 1, ja[1] = 4, ar[1] = 1.0;
ia[2] = 1, ja[2] = 5, ar[2] = 1.0;
ia[3] = 1, ja[3] = 6, ar[3] = 1.0;
glp_set_mat_row(lp, 4, 3, ia, ja, ar);
glp_set_row_bnds(lp, 4, GLP_LO, 19.0, 0.0);
glp_set_row_name(lp, 5, "c5");
ia[1] = 1, ja[1] = 5, ar[1] = 1.0;
ia[2] = 1, ja[2] = 6, ar[2] = 1.0;
ia[3] = 1, ja[3] = 1, ar[3] = 1.0;
glp_set_mat_row(lp, 5, 3, ia, ja, ar);
glp_set_row_bnds(lp, 5, GLP_LO, 31.0, 0.0);
glp_set_row_name(lp, 6, "c6");
ia[1] = 1, ja[1] = 6, ar[1] = 1.0;
ia[2] = 1, ja[2] = 7, ar[2] = 1.0;
ia[3] = 1, ja[3] = 1, ar[3] = 1.0;
glp_set_mat_row(lp, 6, 3, ia, ja, ar);
glp_set_row_bnds(lp, 6, GLP_LO, 45.0, 0.0);
// 设置目标函数
ia[1] = 1, ja[1] = 3, ar[1] = 1.0;
ia[2] = 1, ja[2] = 4, ar[2] = 1.0;
ia[3] = 1, ja[3] = 5, ar[3] = 1.0;
ia[4] = 1, ja[4] = 6, ar[4] = 1.0;
ia[5] = 1, ja[5] = 7, ar[5] = 1.0;
glp_set_obj_coef(lp, 3, 1.0);
glp_set_obj_coef(lp, 4, 1.0);
glp_set_obj_coef(lp, 5, 1.0);
glp_set_obj_coef(lp, 6, 1.0);
glp_set_obj_coef(lp, 7, 1.0);
glp_set_obj_dir(lp, GLP_MIN);
// 求解线性规划问题
glp_simplex(lp, NULL);
ret = glp_get_status(lp);
// 获取结果
if (ret == GLP_OPT)
{
z = glp_get_obj_val(lp);
printf("optimal solution z = %g\n", z);
}
else
{
printf("no optimal solution\n");
}
// 释放内存
glp_delete_prob(lp);
return 0;
}
```
在这个例子中,我们首先创建一个线性规划问题,然后添加变量和约束,设置目标函数,调用glp_simplex函数求解线性规划问题,并获取结果。最后,我们释放内存并结束程序。
需要注意的是,这个线性规划问题是一个混合整数规划问题,但我们没有对变量进行整数约束。如果需要对变量进行整数约束,可以使用GLPK的整数规划求解器。