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494.target-sum.py
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494.target-sum.py
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'''
Description:
Author: liguang-ops
Github: https://github.com/liguang-ops
Date: 2021-09-23 21:07:52
LastEditors: liguang-ops
LastEditTime: 2021-09-24 19:55:44
'''
#
# @lc app=leetcode id=494 lang=python3
#
# [494] Target Sum
#
# @lc code=start
class Solution:
def findTargetSumWays1(self, nums: List[int], target: int) -> int:
"""dfs, 先暴力走一遍,选择就是当前num是加号还是减号,test case 可以通过,
但是submit就会TLE
"""
res = 0
def dfs(ptr):
nonlocal res
if ptr == len(nums):
if sum(nums) == target:
res +=1
return
for sign in [-1, 1]:
nums[ptr] = sign * nums[ptr] #做选择
dfs(ptr + 1) #进入下一层
nums[ptr] = sign * nums[ptr] #撤回选择
dfs(0)
return res
def findTargetSumWays2(self, nums: List[int], target: int) -> int:
"""递归+备忘录
dp(ptr, rest) 指数组在rest下,nums[0...ptr]计算能够的次数
"""
memo = {}
def dp(ptr, rest):
if ptr == len(nums):
if rest == 0:
return 1
return 0
if (ptr, rest) in memo:
return memo[(ptr, rest)]
results = dp(ptr+1, rest-nums[ptr]) + dp(ptr+1, rest + nums[ptr])
memo[(ptr, rest)] = results
return results
return dp(0, target)
def findTargetSumWays3(self, nums: List[int], target: int) -> int:
"""动态规划:
定义dp[i][j]是nums[0...i]能够计算出j的次数
状态转移方程:
首先由两个选择,+ 或者-,应该是取这两种情况的和
dp[i][j] = dp[i-1][j-nums[i]] 注意,这里其实是做+
dp[i][j] = dp[i-1][j-(-nums[i])] ,这里是-
取二者的和
dp[i][j] = dp[i-1][j-nums[i]] + dp[i-1][j-(-nums[i])]
我觉得这样定义是可以的,但是target可以为负数,就无法当作索引了,应该有更好的定义方法
"""
pass
def findTargetSumWays(self, nums: List[int], target: int) -> int:
"""动态规划:转化为背包问题。思路如下:
可以将题目转换一下思路(这样做的前提条件是nums中的数全是非负数):将数组中的数分为两份,一份+背包A,一份减背包B,
sum(A) -sum(B) = target
sum(A) = target + sum(B), 两边再加上sum(A)
sum(A) + sum(A) = target + sum(A) +sum(B) = target + sum(nums)
即sum(A) = (target + sum(nums)) /2,问题转换为nums中有多少个子集A满足左边的等式,即子集背包问题
同样:定义dp[i][j]为当背包容量为j、nums[0...i]有dp[i][j]个子集A恰好能够装满背包
选择很简单,就是nums[i]是否放入背包:
dp[i][j] = dp[i-1][j -nums[i]] 这是放入背包
dp[i][j] = dp[i-1][j], 不放入背包
求和:
dp[i][j] = dp[i-1][j-nums[i-1]] + dp[i-1][j],这里注意nums索引和dp索引差1
base case:
当i=0时,没有东西,肯定无法放满背包, dp[0][j] = 0
当j=0时, 不用放东西,直接就满了,dp[i][0] = 1
"""
if target > sum(nums) or (target + sum(nums)) % 2 !=0 or target + sum(nums) < 0:
return 0
dp = [[0] * ((sum(nums) + target) // 2 + 1) for _ in range(len(nums) + 1)]
for i in range(len(dp)): #初始化状态
dp[i][0] = 1
for i in range(1, len(dp)):
for j in range(0, len(dp[i])): #这里很奇怪,为什么j从0开始。以往的子集背包问题都是从1开始
if j - nums[i-1] >=0: #如果从1开始,测试用例[0,0,0,0,1]\n1不通过。这边不是很明白。
dp[i][j] = dp[i-1][j] + dp[i-1][j-nums[i-1]]
else:
dp[i][j] = dp[i-1][j]
return dp[-1][-1]
# @lc code=end