题目来源：链接: [https://leetcode-cn.com/problems/available-captures-for-rook/].

1.问题描述

在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。

车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。

返回车能够在一次移动中捕获到的卒的数量
示例1：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],
[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],
[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够捕获所有的卒。

示例2：

输入：[[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],
[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],
[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],
[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：0
解释：
象阻止了车捕获任何卒。

示例3：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],
[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],
[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],
[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释： 
车可以捕获位置 b5，d6 和 f5 的卒。

提示:：

1. board.length == board[i].length == 8
2. board[i][j] 可以是 'R'，'.'，'B' 或 'p'
3. 只有一个格子上存在 board[i][j] == 'R'

2.我的解决方案

1. 首先找出 车（R）的坐标 i 和 j
2. 上下左右分别 遍历查找 p ，如果找到B 则break；
3. 如果找到第一个p ，吃掉之后也 break（这个地方需要注意题目不是说吃掉所有的p，而是吃到第一个 p 就停止了）。

代码如下：

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int res = 0;
        int i , j, n;
        for(int l = 0; l < 8; ++l)
        {
            for(int m = 0; m < 8; ++m)
            {
                if(board[l][m] == 'R')
                {
                    i = l;
                    j = m;
                }
            }
        }
        //下面是 上 下 左 右 四个for循环
        for(n = i + 1; n < 8; ++n)
        {
            if(board[n][j] == 'B')
            {
                break;
            }
            if(board[n][j] == 'p')
            {
                ++res;break;
            }
        }
        for(n = i - 1; n >= 0; --n)
        {
            if(board[n][j] == 'B')
            {
                break;
            }
            if(board[n][j] == 'p')
            {
                ++res;break;
            }
        }
        for(n = j + 1; n < 8; ++n)
        {
            if(board[i][n] == 'B')
            {
                break;
            }
            if(board[i][n] == 'p')
            {
                ++res;break;
            }
        }
        for(n = j - 1; n >= 0; --n)
        {
            if(board[i][n] == 'B')
            {
                break;
            }
            if(board[i][n] == 'p')
            {
                ++res;break;
            }
        }
        return res;
    }
};

3.大神们的解决方案

这次没有大神的解决方案？？？
是的，没有。
为啥？？？
别问，问就是我的答案最帅。。。超越了100%。。。哈哈哈

执行用时 : 8 ms, 在Available Captures for Rook的C++提交中击败了100.00% 的用户
内存消耗 : 8.2 MB, 在Available Captures for Rook的C++提交中击败了100.00% 的用户

4.我的收获

vector的find函数

vector <int>::iterator iElement = find(v.begin(), v.end(),3); //查找 容器中元素为3 的返回迭代器的位置

继续加强训练。。。

2019/3/24 胡云层 于南京 34