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Commit 5a0f5b2

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feat: add solutions to lcof2 problem: No.101
lcof2 No.101.Partition Equal Subset Sum
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‎lcof2/剑指 Offer II 101. 分割等和子串/README.md‎

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<p><meta charset="UTF-8" />注意:本题与主站 416&nbsp;题相同:&nbsp;<a href="https://leetcode-cn.com/problems/partition-equal-subset-sum/">https://leetcode-cn.com/problems/partition-equal-subset-sum/</a></p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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题目可以转换为 `0-1` 背包问题,在 m 个数字中选出一些数字(每个数字只能使用一次),这些数字之和恰好等于 `s / 2`(s 表示所有数字之和)。
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也可以用 DFS + 记忆化搜索。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def canPartition(self, nums: List[int]) -> bool:
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s = sum(nums)
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if s % 2 != 0:
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return False
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m, n = len(nums), (s >> 1) + 1
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dp = [[False] * n for _ in range(m)]
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for i in range(m):
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dp[i][0] = True
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if nums[0] < n:
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dp[0][nums[0]] = True
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for i in range(1, m):
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for j in range(n):
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dp[i][j] = dp[i - 1][j]
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if not dp[i][j] and nums[i] <= j:
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dp[i][j] = dp[i - 1][j - nums[i]]
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return dp[-1][-1]
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```
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空间优化:
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```python
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class Solution:
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def canPartition(self, nums: List[int]) -> bool:
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s = sum(nums)
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if s % 2 != 0:
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return False
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m, n = len(nums), (s >> 1) + 1
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dp = [False] * n
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dp[0] = True
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if nums[0] < n:
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dp[nums[0]] = True
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for i in range(1, m):
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for j in range(n - 1, nums[i] - 1, -1):
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dp[j] = dp[j] or dp[j - nums[i]]
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return dp[-1]
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```
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DFS + 记忆化搜索:
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```python
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class Solution:
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def canPartition(self, nums: List[int]) -> bool:
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s = sum(nums)
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if s % 2 != 0:
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return False
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target = s >> 1
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@lru_cache(None)
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def dfs(i, s):
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nonlocal target
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if s > target or i >= len(nums):
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return False
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if s == target:
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return True
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return dfs(i + 1, s) or dfs(i + 1, s + nums[i])
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return dfs(0, 0)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean canPartition(int[] nums) {
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int s = 0;
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for (int x : nums) {
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s += x;
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}
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if (s % 2 != 0) {
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return false;
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}
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int m = nums.length, n = (s >> 1) + 1;
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boolean[] dp = new boolean[n];
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dp[0] = true;
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if (nums[0] < n) {
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dp[nums[0]] = true;
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}
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for (int i = 1; i < m; ++i) {
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for (int j = n - 1; j >= nums[i]; --j) {
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dp[j] = dp[j] || dp[j - nums[i]];
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}
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}
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return dp[n - 1];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool canPartition(vector<int>& nums) {
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int s = 0;
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for (int x : nums) s += x;
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if (s % 2 != 0) return false;
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int m = nums.size(), n = (s >> 1) + 1;
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vector<bool> dp(n);
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dp[0] = true;
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if (nums[0] < n) dp[nums[0]] = true;
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for (int i = 1; i < m; ++i)
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{
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for (int j = n - 1; j >= nums[i]; --j)
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{
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dp[j] = dp[j] || dp[j - nums[i]];
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}
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}
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return dp[n - 1];
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}
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};
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```
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### **Go**
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```go
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func canPartition(nums []int) bool {
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s := 0
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for _, x := range nums {
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s += x
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}
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if s%2 != 0 {
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return false
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}
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m, n := len(nums), (s>>1)+1
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dp := make([]bool, n)
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dp[0] = true
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if nums[0] < n {
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dp[nums[0]] = true
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}
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for i := 1; i < m; i++ {
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for j := n - 1; j >= nums[i]; j-- {
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dp[j] = dp[j] || dp[j-nums[i]]
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}
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}
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return dp[n-1]
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}
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```
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### **...**
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class Solution {
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public:
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bool canPartition(vector<int>& nums) {
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int s = 0;
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for (int x : nums) s += x;
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if (s % 2 != 0) return false;
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int m = nums.size(), n = (s >> 1) + 1;
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vector<bool> dp(n);
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dp[0] = true;
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if (nums[0] < n) dp[nums[0]] = true;
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for (int i = 1; i < m; ++i)
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{
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for (int j = n - 1; j >= nums[i]; --j)
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{
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dp[j] = dp[j] || dp[j - nums[i]];
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}
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}
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return dp[n - 1];
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}
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};
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func canPartition(nums []int) bool {
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s := 0
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for _, x := range nums {
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s += x
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}
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if s%2 != 0 {
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return false
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}
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m, n := len(nums), (s>>1)+1
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dp := make([]bool, n)
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dp[0] = true
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if nums[0] < n {
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dp[nums[0]] = true
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}
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for i := 1; i < m; i++ {
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for j := n - 1; j >= nums[i]; j-- {
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dp[j] = dp[j] || dp[j-nums[i]]
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}
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}
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return dp[n-1]
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}
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class Solution {
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public boolean canPartition(int[] nums) {
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int s = 0;
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for (int x : nums) {
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s += x;
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}
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if (s % 2 != 0) {
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return false;
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}
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int m = nums.length, n = (s >> 1) + 1;
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boolean[] dp = new boolean[n];
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dp[0] = true;
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if (nums[0] < n) {
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dp[nums[0]] = true;
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}
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for (int i = 1; i < m; ++i) {
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for (int j = n - 1; j >= nums[i]; --j) {
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dp[j] = dp[j] || dp[j - nums[i]];
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}
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}
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return dp[n - 1];
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}
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}
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class Solution:
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def canPartition(self, nums: List[int]) -> bool:
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s = sum(nums)
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if s % 2 != 0:
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return False
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m, n = len(nums), (s >> 1) + 1
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dp = [False] * n
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dp[0] = True
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if nums[0] < n:
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dp[nums[0]] = True
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for i in range(1, m):
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for j in range(n - 1, nums[i] - 1, -1):
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dp[j] = dp[j] or dp[j - nums[i]]
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return dp[-1]

‎solution/0400-0499/0416.Partition Equal Subset Sum/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -40,6 +40,8 @@
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题目可以转换为 `0-1` 背包问题,在 m 个数字中选出一些数字(每个数字只能使用一次),这些数字之和恰好等于 `s / 2`(s 表示所有数字之和)。
4242

43+
也可以用 DFS + 记忆化搜索。
44+
4345
<!-- tabs:start -->
4446

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### **Python3**
@@ -89,6 +91,28 @@ class Solution:
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return dp[-1]
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```
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DFS + 记忆化搜索:
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```python
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class Solution:
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def canPartition(self, nums: List[int]) -> bool:
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s = sum(nums)
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if s % 2 != 0:
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return False
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target = s >> 1
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@lru_cache(None)
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def dfs(i, s):
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nonlocal target
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if s > target or i >= len(nums):
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return False
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if s == target:
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return True
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return dfs(i + 1, s) or dfs(i + 1, s + nums[i])
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return dfs(0, 0)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->

‎solution/0400-0499/0416.Partition Equal Subset Sum/README_EN.md‎

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return dp[-1]
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```
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```python
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class Solution:
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def canPartition(self, nums: List[int]) -> bool:
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s = sum(nums)
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if s % 2 != 0:
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return False
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target = s >> 1
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@lru_cache(None)
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def dfs(i, s):
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nonlocal target
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if s > target or i >= len(nums):
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return False
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if s == target:
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return True
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return dfs(i + 1, s) or dfs(i + 1, s + nums[i])
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return dfs(0, 0)
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```
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### **Java**
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```java

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