动态规划之博弈问题中的斜遍历数组

class Pair {    int fir, sec;    Pair(int fir, int sec) {        this.fir = fir;        this.sec = sec;    }}
/* 返回游戏最后先手和后手的得分之差 */int stoneGame(int[] piles) {    int n = piles.length;    // 初始化 dp 数组    Pair[][] dp = new Pair[n][n];    for (int i = 0; i < n; i++)         for (int j = i; j < n; j++)            dp[i][j] = new Pair(0, 0);    // 填入 base case    for (int i = 0; i < n; i++) {        dp[i][i].fir = piles[i];        dp[i][i].sec = 0;    }    // 斜着遍历数组    for (int l = 2; l <= n; l++) {        for (int i = 0; i <= n - l; i++) {            int j = l + i - 1;            // 先手选择最左边或最右边的分数            int left = piles[i] + dp[i+1][j].sec;            int right = piles[j] + dp[i][j-1].sec;            // 套用状态转移方程            if (left > right) {                dp[i][j].fir = left;                dp[i][j].sec = dp[i+1][j].fir;            } else {                dp[i][j].fir = right;                dp[i][j].sec = dp[i][j-1].fir;            }        }    }    Pair res = dp[0][n-1];    return res.fir - res.sec;}