n件物品和最大承受重量为w的背包，其中第i件物品的重量是weight[i]，得到的价值是value[i]，每件物品只能用一次，求解将哪些物品装入背包里物品价值总和最大。
1. 确定dp数组以及下标的含义
dp[i][j] 表示从下标为[0-i]的物品里任意取，放进容量为j的背包，价值总和最大是多少。i：物品编号，j：背包容量，dp[i][j]：价值总和
2. 确定递推公式
递归公式： dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
3. dp数组初始化
dp[i][0]：背包容量为0因此价值总和为0
dp[0][j]：存放物品0的时候背包价值总和，当j<weight[0]价值总和为0，其余情况价值总和为value[0]
4. 确定遍历顺序：先遍历物品/背包重量
5. 举例推导dp数组
class Solution(object):
def canPartition(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
if sum(nums) % 2: return False
# 要求找到元素和为sum/2的子集
target = sum(nums) // 2
# dp[i][j]代表可装物品为0-i，背包容量为j的情况下，背包内容量的最大价值（最大子集和）
dp = [[0 for _ in range(target+1)] for _ in range(len(nums))]
# 背包容量为0且只能放入nums[i]的情况下背包最大价值（最大子集和）
for j in range(nums[0],target+1):
dp[0][j] = nums[0]
# 递推公式（每行从左到右遍历）
for i in range(1, len(nums)):
for j in range(1, target+1):
# 当背包不能容纳nums[i]时
if j < nums[i]:
dp[i][j] = dp[i-1][j]
# 当背包可以容纳nums[i]时
else:
dp[i][j] = max(dp[i-1][j], dp[i-1][j-nums[i]] + nums[i])
return dp[-1][-1] == target使用滑动数组：
class Solution(object):
def canPartition(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
if not nums or sum(nums) % 2: return False
target = sum(nums) // 2
# 滑动数组（因为只需要保留上一行的值因此可以使用一维数组）
dp = [0 for _ in range(target+1)]
for i in range(len(nums)):
for j in range(target, nums[i]-1, -1):
dp[j] = max(dp[j], dp[j-nums[i]] + nums[i])
return dp[-1] == target
class Solution(object):
def lastStoneWeightII(self, stones):
"""
:type stones: List[int]
:rtype: int
"""
# 将stones分为重量差不多的两堆
target = sum(stones) // 2
# dp设为滑动数组，dp[j]存储最接近重量j其不超过的石头总重量
dp = [0 for _ in range(target+1)]
# 遍历stones
for i in range(len(stones)):
for j in range(target, stones[i]-1, -1):
dp[j] = max(dp[j], dp[j-stones[i]]+stones[i])
return sum(stones)-2*dp[-1]
class Solution(object):
def findTargetSumWays(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if len(nums) == 1 and nums[0] != target and nums[0] != -1 * target:
return 0
sum_nums = sum(nums)
if target > sum_nums or target < -1 * sum_nums: return 0
# dp[][sum_nums]为中心点
dp = [[0 for _ in range(2*sum_nums+1)] for _ in range(len(nums))]
for j in range(2*sum_nums+1):
if abs(j - sum_nums) == nums[0]:
if nums[0] == 0: dp[0][j] = 2
else: dp[0][j] = 1
for i in range(len(nums)-1):
for j in range(2*sum_nums+1):
if dp[i][j] == 0: continue
dp[i + 1][j - nums[i + 1]] += dp[i][j]
dp[i + 1][j + nums[i + 1]] += dp[i][j]
return dp[-1][target+sum_nums]第二种方法：
假设nums = [3,1,2,5,4]，此时选3和1前面的符号为加号，其余为减号，那么此时加法获得的和为3+1=4，减法获得的和为2+5+4=11，也可以通过sum(nums)-3-1=15-3-1=11获得。以此类推，设加法获得的和为x，则减法获得的和为(sum-x)，问题转化满足x-(sum-x)=target / x=(sum+target)/2有几种方案。
class Solution(object):
def findTargetSumWays(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
sum_nums = sum(nums)
# 注意边界条件
if abs(target) > sum_nums or (sum_nums + target) % 2 == 1: return 0
bagSize = (sum_nums + target) // 2
dp = [0] * (bagSize + 1)
dp[0] = 1
for i in range(len(nums)):
for j in range(bagSize, nums[i] - 1, -1):
dp[j] += dp[j - nums[i]]
return dp[bagSize]
dp[i][j]：最多有i个0和j个1的strs的最大子集的大小为dp[i][j]。
class Solution(object):
def findMaxForm(self, strs, m, n):
"""
:type strs: List[str]
:type m: int
:type n: int
:rtype: int
"""
# dp[i][j]：最多有i个0和j个1的strs的最大子集的大小为dp[i][j]
dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
for str in strs:
onenum = str.count('1')
zeronum = str.count('0')
for i in range(m, zeronum-1, -1):
for j in range(n, onenum-1, -1):
dp[i][j] = max(dp[i][j], dp[i-zeronum][j-onenum] + 1)
return dp[m][n]
class Solution(object):
def change(self, amount, coins):
"""
:type amount: int
:type coins: List[int]
:rtype: int
"""
# dp[i][j]表示coins[i]的面值加入排列后，总和为amount的组合数
dp = [0 for _ in range(amount+1)]
dp[0] = 1
for i in range(len(coins)):
for j in range(coins[i], amount+1):
dp[j] += dp[j-coins[i]]
return dp[-1]
class Solution(object):
def combinationSum4(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
dp = [0 for _ in range(target+1)]
dp[0] = 1
for j in range(target+1):
for i in range(len(nums)):
if j-nums[i] >= 0:
dp[j] += dp[j-nums[i]]
return dp[-1]
class Solution(object):
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
dp = [0 for _ in range(n+1)]
dp[0] = 1
for j in range(n+1):
for step in range(1,3):
if j - step >= 0:
dp[j] += dp[j-step]
return dp[-1]
class Solution(object):
def coinChange(self, coins, amount):
"""
:type coins: List[int]
:type amount: int
:rtype: int
"""
dp = [amount + 1] * (amount + 1)
dp[0] = 0
for coin in coins:
for j in range(coin, amount + 1):
dp[j] = min(dp[j], dp[j - coin] + 1)
if dp[amount] < amount + 1:
return dp[amount]
else:
return -1
class Solution(object):
def numSquares(self, n):
"""
:type n: int
:rtype: int
"""
square, i = [], 1
while i**2 <= n:
square.append(i**2)
i += 1
dp = [n+1] * (n+1)
dp[0] = 0
for i in range(len(square)):
for j in range(square[i], n+1):
dp[j] = min(dp[j], dp[j-square[i]]+1)
if dp[n] > n+1:
return -1
else:
return dp[n]
class Solution(object):
def wordBreak(self, s, wordDict):
"""
:type s: str
:type wordDict: List[str]
:rtype: bool
"""
# dp[j]=1即截至此处s[:j+1]可划分，dp[j]=0即不可划分
dp = [0 for _ in range(len(s)+1)]
dp[0] = 1
for j in range(1, len(s)+1):
for word in wordDict:
if j >= len(word):
dp[j] = dp[j] or (dp[j - len(word)] and word == s[j - len(word):j])
return dp[-1] == 1