原题链接在这里：http://www.lintcode.com/en/problem/coins-in-a-line-iii/
题目：
There are n coins in a line. Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Could you please decide the first player will win or lose?
Given array A = [3,2,2], return true.
Given array A = [1,2,4], return true.
Given array A = [1,20,4], return false.
题解：
DP问题. 状态dp[l][r]第i到第j的硬币时，先手去硬币的人最后最多去硬币价值.
转移方程, pickLeft = min(dp[l+2][r], dp[l+1][r-1])+values[l]
pickRight = min(dp[l][r-2], dp[l-1][r-1])+values[r]
dp[l][r] = max(pickLeft, pickRight)
初始化l == r时, dp[l][r] = values[l]
l+1 == r时, dp[l][r] = max(values[l], values[l+1])
答案dp[0][values.length-1] > sum/2
Time Complexity: O(n). Space: O(n^2).
AC Java:
1 public class Solution { 2 /** 3 * @param values: an array of integers 4 * @return: a boolean which equals to true if the first player will win 5 */ 6 public boolean firstWillWin(int[] values) { 7 if(values == null || values.length == 0){ 8 return false; 9 } 10 11 int n = values.length; 12 int [][] dp = new int[n][n]; 13 boolean [][] used = new boolean[n][n]; 14 int sum = 0; 15 for(int val : values){ 16 sum += val; 17 } 18 return sum < 2*memorySearch(values, 0, values.length-1, dp, used); 19 } 20 21 private int memorySearch(int [] values, int l, int r, int [][] dp, boolean [][] used){ 22 if(used[l][r]){ 23 return dp[l][r]; 24 } 25 used[l][r] = true; 26 if(l>r){ 27 dp[l][r] = 0; 28 }else if(l == r){ 29 dp[l][r] = values[l]; 30 }else if(l + 1 == r){ 31 dp[l][r] = Math.max(values[l], values[l+1]); 32 }else{ 33 int pickLeft = Math.min(memorySearch(values, l+2, r, dp, used), memorySearch(values, l+1, r-1, dp, used)) + values[l]; 34 int pickRight = Math.min(memorySearch(values, l+1, r-1, dp, used), memorySearch(values, l, r-2, dp, used)) + values[r]; 35 dp[l][r] = Math.max(pickLeft, pickRight); 36 } 37 return dp[l][r]; 38 } 39 }
原文：http://www.cnblogs.com/Dylan-Java-NYC/p/6410366.html