# Coins In A Line Iii

## Coins In A Line Iii

## 题目地址

<https://www.lintcode.com/problem/coins-in-a-line/> <https://zhengyang2015.gitbooks.io/lintcode/coins_in_a_line_iii_396.html>

## 题目描述

```
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.
```

## 代码

### Approach #1 DP

```java
public class Solution {
  public boolean firstWillWin(int[] values) {
    int n = values.length;
    int[] sum = new int[n + 1];
    sum[0] = 0;

    for (int i = 1; i <= n; i++) /* Get the sum in ith index */
      sum[i] = sum[i - 1] + values[i - 1];

    int[][] dp = new int[n][n];
    for (int i = 0; i < n; i++) 
      dp[i][i] = values[i]; /* Treat these like the leaf nodes */

    for (int len = 2; len <= n; len++) {
      for (int i = 0; i < n; i++) {
        int j = i + len - 1;
        if (j >= n) continue;
        int s = sum[j + 1] - sum[i];
        dp[i][j] = Math.max(s - dp[i + 1][j], s - dp[i][j - 1]);
      }
    }

    return dp[0][n - 1] > sum[n] / 2;
  }
}
```

### Approach #2 DFS

`dp[start][end] => [values[start-1], values[end-1]]`

```java
public class Solution {
    /**
     * @param values: an array of integers
     * @return: a boolean which equals to true if the first player will win
     */
    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];
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/dynamic-programming/qu-jian-xing/coins-in-a-line-iii.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.