https://zhengyang2015.gitbooks.io/lintcode/content/coins_in_a_line_iii_396.html
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?
Example
Given array A = [3,2,2], return true.
Given array A = [1,2,4], return true.
Given array A = [1,20,4], return false.
Challenge
Follow Up Question:
If n is even. Is there any hacky algorithm that can decide whether first player will win or lose in O(1) memory and O(n) time?
用memory search的方法解，和II类似。dp[i][j]表示在i－j区间里我们能取到的最大值。
我们每次能取的的值为开头或者结尾元素，因此我们在i－j区间能取的最大值为：i－j区间元素的总和－对手在剩下区间能取到的元素的总和的较小值。因此，状态函数为：
dp[i][j] = sum[i][j] - min(dp[i+1][j], dp[i][j-1])。
DFS+Cache
public boolean firstWillWin(int[] values) {
// write your code here
if(values == null || values.length == 0){
return false;
}
int n = values.length;
int[] sum = new int[n + 1];
sum[0] = 0;
for(int i = 1; i <= n; i++){
sum[i] = sum[i - 1] + values[i - 1];
}
int[][] dp = new int[n + 1][n + 1];
boolean[][] visit = new boolean[n + 1][n + 1];
return search(1, n, sum, dp, visit) > sum[n] / 2;
}
private int search(int start, int end, int[] sum, int[][] dp, boolean[][] visit){
if(visit[start][end]){
return dp[start][end];
}
if(start == end){
visit[start][end] = true;
return dp[start][end] = sum[end] - sum[start - 1];
}
int max = (sum[end] - sum[start - 1]) - Math.min(search(start, end - 1, sum, dp, visit), search(start + 1, end, sum, dp, visit));
visit[start][end] = true;
dp[start][end] = max;
return dp[start][end];
}
dp题，主要是要推导出公式：
dp[i][j]=max(A[i]+sum[i+1][j]-dp[i+1][j], A[j]+sum[i][j-1]-dp[i][j-1])
A[i]+sum[i+1][j] 和 A[j]+sum[i][j-1]都等于sum[i][j],因此最后公式成为：
dp[i][j]=sum[i][j]-min(dp[i+1][j],dp[i][j-1])
比较特别的dp题。game theory。
dp[i][j]=max(A[i]+sum[i+1][j]-dp[i+1][j], A[j]+sum[i][j-1]-dp[i][j-1])
A[i]+sum[i+1][j] 和 A[j]+sum[i][j-1]都等于sum[i][j],因此最后公式成为：
dp[i][j]=sum[i][j]-min(dp[i+1][j],dp[i][j-1])
比较特别的dp题。game theory。
public boolean firstWillWin(int[] values) {
int len = values.length;
if (len <= 1) {
return true;
}
int[][] store = new int[len][len];
int[][] sum = new int[len][len];
for (int i = 0; i < len; i++) {
for (int j = i; j < len; j++) {
sum[i][j] = i == j ? values[j] : sum[i][j-1] + values[j];
}
}
for (int i = len - 1; i >= 0; i--) {
for (int j = i; j < len; j++) {
if (i == j) {
store[i][j] = values[i];
} else {
int cur = Math.min(store[i+1][j], store[i][j-1]);
store[i][j] = sum[i][j] - cur;
}
}
}
return store[0][len - 1] > sum[0][len-1] - store[0][len - 1];
}
备忘录，dp[left][right]表示从left到right所能取到的最大值，因为双方都取最优策略，所以取完一个后，对手有两种选择，我们要加让对手得到更多value的那种方案，也就是自己得到更少value的方案。
