2019-03-17 Max Area of Island

首先最简单的思路就是深度优先遍历：

class Solution {

public:

    const int CARRY = 51;

    int maxAreaOfIsland(vector >& grid) {

        map earth;

        for (int i = 0; i < grid.size(); i++) {

            for (int j = 0; j < grid[0].size(); j++) {

if (grid[i][j] == 1) {

                earth.insert(pair(i*CARRY+j, false));

}

            }

        }

        int theMax = 0;

        for (auto one : earth) {

            if (!one.second) {

                int area = travel(one.first, earth);

                theMax = std::max(theMax, area);

            }

        }

        return theMax;

    }

    int travel(int pos, map &earth) {

        int area = 0;

        if (earth.find(pos) != earth.end()) {

            if (earth[pos] == false) {

                earth[pos] = true; 

                int x = pos / CARRY, y = pos % CARRY;

                area = 1

                    + travel((x-1)*CARRY + y, earth)

                    + travel((x+1)*CARRY + y, earth)

                    + travel(x*CARRY + y - 1, earth)

                    + travel(x*CARRY + y + 1, earth);

            }

        }

        return area;

    }

};

结果：

Runtime:52 ms, faster than 12.74% of C++ online submissions for Max Area of Island.

Memory Usage:20.5 MB, less than 7.80% of C++ online submissions for Max Area of Island.

方案二：

一行一行遍历，每一次查看有没有左边或者上边相邻的点，有则连在一起构成新的岛屿。

class Solution {

public:

    const int CARRY = 51;

    map > > earthMap;

    int maxAreaOfIsland(vector >& grid) {

        int theMax = 0;

        for (int i = 0; i < grid.size(); i++) {

            for (int j = 0; j < grid[0].size(); j++) {

                if (grid[i][j] == 1) {

                    int area = enlargeIsland(posInt(i,j));

                    theMax = std::max(theMax, area);

                }

            }

        }

        return theMax;

    }

    int enlargeIsland(int pos) {

        int i = pos % CARRY, j = pos / CARRY;

        list > > nearByEarths;

        if (earthMap.find(posInt(i - 1, j)) != earthMap.end()) {

            nearByEarths.push_back(earthMap[posInt(i - 1, j)]);

earthMap.erase(posInt(i - 1, j));

        }

if (earthMap.find(posInt(i, j - 1)) != earthMap.end()) {

nearByEarths.push_back(earthMap[posInt(i, j - 1)]);

earthMap.erase(posInt(i, j - 1));

        }

        shared_ptr > joinedEarth = make_shared >();

earthMap.insert(pair > >(posInt(i, j), joinedEarth));

joinedEarth->insert(posInt(i, j));

        int joinedArea = 1;

        for (auto earthIter : nearByEarths) {

            for (int earthPos : *earthIter) {

                earthMap[earthPos] = joinedEarth;

                if (joinedEarth->find(earthPos) == joinedEarth->end()) {

joinedArea++;

joinedEarth->insert(earthPos);

}

            }

        }

        return joinedArea;

    }

    int inline posInt(int x, int y) {

        return x + CARRY * y;

    }

};

本来还想着方案二回快些，执行结果发现还超时了……

2-1、还得思考下为什么方案二比一慢

2-2、有没有别的方案，因为目前方案一在leetcode的排名看来，还不是最快的算法。

方案三：

看了下排名靠前的答案，跟我方案一的思路差不多呀，只不过我的办法啰嗦了点？

因此作以下改进：

class Solution {

public:   

    int maxAreaOfIsland(vector >& grid) {

        int theMax = 0;

        for (int i = 0; i < grid.size(); i++) {

            for (int j = 0; j < grid[0].size(); j++) {

if (grid[i][j] == 1) {

                int area = travel(i, j, grid);

                theMax = std::max(theMax, area);

}

            }

        }

        return theMax;

    }

    int travel(int x, int y, vector > &grid) {

        int area = 0;

        if (grid[x][y] == 1) {

grid[x][y] = 0; 

area = 1

+ (x - 1 >= 0 ? travel(x-1, y, grid) : 0)

+ (x + 1 < grid.size() ? travel(x+1, y, grid) : 0)

+ (y - 1 >= 0 ? travel(x,  y - 1, grid) : 0)

+ (y + 1 < grid[0].size() ? travel(x, y + 1, grid) : 0);

        }

        return area;

    }

};

成绩：

Runtime:20 ms, faster than 98.60% of C++ online submissions for Max Area of Island.

Memory Usage:14.7 MB, less than 52.48% of C++ online submissions for Max Area of Island.