已知一个数字列表nums和一个目标数字target，请编写一个函数，从左往右，找出其中任意两个连续且和值刚好等于target的数字，一旦找到，需要从移除其中的第二个数字。以此类推，最终返回处理后的列表。
示例：
输入：nums=[1, 2, 3, 4, 5]，target=3。
返回：[1, 3, 4, 5]。
解释：因为1+2=3，所以移除2；最终剩下[1,3,4,5]。
题目难度：简单
题目来源：CodeWars：Double Trouble
def solution(nums: list, target: int) -> list:
# your code
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
def solution(nums: list, target) -> list:
new_list = []
for i in range(1, len(nums)):
for j in range(1, i + 1):
if nums[i - j]:
pre_n = nums[i - j]
break
else:
pre_n = None
if pre_n and nums[i] + pre_n == target:
nums[i] = None
else:
new_list.append(nums[i])
return nums[0:1] + new_list
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
很笨的方法
def solution(nums: list,target) -> list:
l=[]
for i in range(len(nums)):
l.append(nums[i])
if len(l)>1:
if l[-1] + l[-2] == target:
l.pop()
return l
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
def solution(nums_list:list, target_number):
num1 = nums_list[0]
res_list = [num1,]
for i,v in enumerate(nums_list):
if i>0:
num2 = v
if (num1+num2) == target_number:
continue
else:
num1 = nums_list[i]
res_list.append(num2)
return res_list
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
def solution(nums:list,target):
i=0
while True:
sum1=nums[i]+nums[i+1]
# 判断连续两个值是否等于目标值
if sum1==target:
# 相等的话就删除该值
nums.pop(i+1)
# 判断列表的长度-1是否与循环次数相等，相等则整个列表已经遍历一遍，终止循环
if i==len(nums)-1:
break
# 跳出当前循环，继续下一次循环
continue
# 连续两个值不等于目标值
else:
i+=1
# 判断列表的长度-1是否与循环次数相等，相等则整个列表已经遍历一遍，终止循环
if i==len(nums)-1:
break
return nums
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
def solution(nums: list, target) -> list:
length = len(nums)
i = 0
while i < length - 1:
if nums[i] + nums[i + 1] == target:
nums.pop(i + 1)
length = length - 1
else:
i += 1
return nums
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
def solution(nums: list, target: int) -> list:
# your code
index = 0
while index < len(nums)-1:
if nums[index] + nums[index+1] == target:
nums.pop(index+1)
continue
else:
index += 1
return nums
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
public List<Integer> solution(List<Integer> list, int target) {
int i = 0;
while (i < list.size() - 1) {
if (list.get(i) + list.get(i + 1) == target) {
list.remove(i + 1);
continue;
} else {
i++;
}
}
return list;
}
def double_trouble(nums:list,target:int)->list:
left = 0
right = 1
new_nums = [nums[left]]
while right< len(nums):
if nums[left]+nums[right] == target:
right+=1
else:
new_nums.append(nums[right])
left= right
right+=1
return new_nums后面咱们做个系统，让大家自动练习冲榜吧
def solution(nums: list,target) -> list:
i = 0
while i < len(nums)-1:
sum1 = nums[i]+nums[i+1]
if sum1 == target:
nums.pop(i+1)
continue
else:
i = i+1
return nums
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
def solution(nums: list, target: int) -> list:
newlist = [nums[0]]
for i in range(1,len(nums),1):
if nums[i] + newlist[len(newlist)-1] != target:
newlist.append(nums[i])
return newlist
def solution(nums: list, num: int) -> list:
# your code
for j in range(len(nums)):
for i in range(len(nums) - 1):
n = nums[i] + nums[i + 1]
if n == num:
del nums[i + 1]
break
return nums
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]public static List<Integer> solution(List<Integer> lis, int num){
int delIndex = 0;
for (int i = 0; i < lis.size()-1; i++) {
int a = lis.get(i)+lis.get(i+1);
if (num==a){
delIndex=i+1;
break;
}
}
if (delIndex==0){
System.out.println("未找到需要删除元素");
return lis;
}
lis.remove(delIndex);
return lis;
}
def solution(nums: list, target: int) -> list:
i = 0
while True:
if nums[i] + nums[i + 1] == target:
nums.pop(i + 1)
if i == len(nums) - 1:
break
else:
continue
else:
i += 1
if i == len(nums) - 1:
break
return nums
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]
好的 
参考题解
def solution(nums: list, target: int) -> list:
result = [nums[0]]
i, j = 0, 0
for idx in range(len(nums)-1):
j+=1
if nums[i]+nums[j] == target:
continue
else:
result.append(nums[j])
i=j
return result
assert solution([1, 3, 5, 6, 7, 4, 3], 7) == [1, 3, 5, 6, 7, 4]
assert solution([4, 1, 1, 1, 4], 2) == [4, 1, 4]
assert solution([2, 6, 2], 8) == [2, 2]
assert solution([2, 2, 2, 2, 2, 2], 4) == [2]