LeetCode Predict the Winner
原题链接在这里：https://leetcode.com/problems/predict-the-winner/description/
题目：
Given an array of scores that are non-negative integers. Player 1 picks one of the numbers from either end of the array followed by the player 2 and then player 1 and so on. Each time a player picks a number, that number will not be available for the next player. This continues until all the scores have been chosen. The player with the maximum score wins.
Given an array of scores, predict whether player 1 is the winner. You can assume each player plays to maximize his score.
Example 1:
Input: [1, 5, 2]
Output: False
Explanation: Initially, player 1 can choose between 1 and 2.
If he chooses 2 (or 1), then player 2 can choose from 1 (or 2) and 5. If player 2 chooses 5, then player 1 will be left with 1 (or 2).
So, final score of player 1 is 1 + 2 = 3, and player 2 is 5.
Hence, player 1 will never be the winner and you need to return False.
Example 2:
Input: [1, 5, 233, 7]
Output: True
Explanation: Player 1 first chooses 1. Then player 2 have to choose between 5 and 7. No matter which number player 2 choose, player 1 can choose 233.
Finally, player 1 has more score (234) than player 2 (12), so you need to return True representing player1 can win.
Note:
- 1 <= length of the array <= 20.
- Any scores in the given array are non-negative integers and will not exceed 10,000,000.
- If the scores of both players are equal, then player 1 is still the winner.
题解：
dp[i][j]是nums 从i到j这一段[i, j] 先手的player 比 后手多得到多少分.
先手 pick first. 递推时 dp[i][j] = Math.max(nums[i]-dp[i+1][j], nums[j]-dp[i][j-1]). 如果A选了index i的score, B只能选择[i+1, j]区间内的score. 如果A选了index j的score, B只能选择[i, j-1]区间内的score.
看到计算dp[i][j]时, i 需要 i+1, j 需要 j-1. 所以循环时 i从大到小, j 从小到大.
初始化区间内只有一个数字时就是能得到的最大分数.
答案看[0, nums.length-1]区间内 A得到的score是否大于等于0.
Time Complexity: O(len^2). len = nums.length.
Space: O(len^2).
AC Java:
class Solution {
public boolean PredictTheWinner(int[] nums) {
if(nums == null || nums.length == 0){
return true;
}
int len = nums.length;
int [][] dp = new int[len][len];
for(int i = len-1; i>=0; i--){
for(int j = i+1; j<len; j++){
int head = nums[i]-dp[i+1][j];
int tail = nums[j]-dp[i][j-1];
dp[i][j] = Math.max(head, tail);
}
}
return dp[0][len-1] >= 0;
}
}
空间优化.
Time Complexity: O(len^2). len = nums.length.
Space: O(len).
AC Java:
class Solution {
public boolean PredictTheWinner(int[] nums) {
if(nums == null || nums.length == 0){
return true;
}
int len = nums.length;
int [] dp = new int[len];
for(int i = len-1; i>=0; i--){
for(int j = i+1; j<len; j++){
int head = nums[i]-dp[j];
int tail = nums[j]-dp[j-1];
dp[j] = Math.max(head, tail);
}
}
return dp[len-1] >= 0;
}
}
另一种implementation.
Time Complexity: O(len^2). len = nums.length.
Space: O(len^2).
class Solution {
public boolean PredictTheWinner(int[] nums) {
if(nums == null || nums.length == 0){
return true;
}
int n = nums.length;
int [][] dp = new int[n][n];
for(int i = 0; i<n; i++){
dp[i][i] = nums[i];
}
for(int size = 1; size<n; size++){
for(int i = 0; i+size<n; i++){
dp[i][i+size] = Math.max(nums[i]-dp[i+1][i+size], nums[i+size]-dp[i][i+size-1]);
}
}
return dp[0][n-1] >= 0;
}
}
Exact the same as Stone Game.
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