# Stone Game IV (H)

## 题目

Alice and Bob take turns playing a game, with Alice starting first.

Initially, there are `n` stones in a pile. On each player's turn, that player makes a move consisting of removing any non-zero square number of stones in the pile.

Also, if a player cannot make a move, he/she loses the game.

Given a positive integer `n`. Return `True` if and only if Alice wins the game otherwise return `False`, assuming both players play optimally.

Example 1:

``````Input: n = 1
Output: true
Explanation: Alice can remove 1 stone winning the game because Bob doesn't have any moves.
``````

Example 2:

``````Input: n = 2
Output: false
Explanation: Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0).
``````

Example 3:

``````Input: n = 4
Output: true
Explanation: n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0).
``````

Example 4:

``````Input: n = 7
Output: false
Explanation: Alice can't win the game if Bob plays optimally.
If Alice starts removing 4 stones, Bob will remove 1 stone then Alice should remove only 1 stone and finally Bob removes the last one (7 -> 3 -> 2 -> 1 -> 0).
If Alice starts removing 1 stone, Bob will remove 4 stones then Alice only can remove 1 stone and finally Bob removes the last one (7 -> 6 -> 2 -> 1 -> 0).
``````

Example 5:

``````Input: n = 17
Output: false
Explanation: Alice can't win the game if Bob plays optimally.
``````

Constraints:

• `1 <= n <= 10^5`

## 代码实现

### Java

#### 记忆化搜索

``````class Solution {
public boolean winnerSquareGame(int n) {
Map<Integer, Boolean> record = new HashMap<>();
return dfs(n, record);
}

private boolean dfs(int n, Map<Integer, Boolean> record) {
if (n == 0) {
return false;
}

if (record.containsKey(n)) {
return record.get(n);
}

for (int i = 1; i*i <= n;i++) {
if (!dfs(n - i * i, record)) {
record.put(n, true);
return true;
}
}

record.put(n, false);
return false;
}
}
``````

#### 动态规划

``````class Solution {
public boolean winnerSquareGame(int n) {
boolean[] dp = new boolean[n + 1];
for (int i = 1; i <= n; i++) {
for (int j = 1; j * j <= i; j++) {
if (!dp[i - j * j]) {
dp[i] = true;
break;
}
}
}
return dp[n];
}
}
``````