Stone Game IV (H)
題目
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
題意
兩人輪流取數，每次只能取走一個平方數，當前回合無法取出平方數的人算輸，在都采取最優策略的前提下，判斷先取的人是否一定贏，
思路
假設f(n)=true表示總數為n時先取的人能贏，若取走的平方數為k，很明顯只要f(n-k)為false(第二個取的人會輸)，第一個取的人就能贏，兩種實作方式：記憶化搜索和動態規劃，
代碼實作
Java
記憶化搜索
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;
}
}
動態規劃
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];
}
}
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