leetcode 200.岛屿问题 深搜、宽搜、并查集

1.深搜

func numIslands(_ grid: [[Character]]) -> Int {
        var visit = Array.init(repeating: Array.init(repeating: false, count: grid[0].count), count: grid.count)
        var count = 0
        for i in 0..=0 && newX=0 && newY

2.宽搜

func numIslands(_ grid: [[Character]]) -> Int {
        var visit = Array.init(repeating: Array.init(repeating: false, count: grid[0].count), count: grid.count)
        var count = 0
        var queue = [[Int]]()
        
        for i in 0..=0 && newX=0 && newY

3. 并查集

class UnionFind: NSObject {
        var parent:[Int]!
        var count = 0
        var rank:[Int]!
        
        public init(_ grid:[[Character]]) {
            let m = grid.count
            let n = grid[0].count
            
            parent = Array.init(repeating: 0, count: m*n)
            rank = Array.init(repeating: 0, count: m*n)

            for i in 0.. Int {
            var x1 = x
            while x1 != parent[x1] {
                x1 = parent[x1]
            }
            return x1
        }
        
        func union(_ x:Int,_ y:Int) {
            let rootx = find(x)
            let rooty = find(y)
            
            if rootx != rooty {
                if rank[rootx] > rank[rooty] {
                    parent[rooty] = rootx
                }else if rank[rootx] < rank[rooty]{
                    parent[rootx] = rooty
                }else{
                    parent[rooty] = rootx
                    rank[rootx] += 1
                }
                count -= 1
            }
        }
    }
    
    func numIslands(_ oldGrid: [[Character]]) -> Int {
        if oldGrid.count == 0 || oldGrid[0].count == 0 {
            return 0
        }
        var grid = oldGrid
        
        let nr = grid.count
        let nc = grid[0].count
        
        let uf = UnionFind.init(grid)
        
        let directions = [[-1,0],[1,0],[0,1],[0,-1]]
        
        for i in 0..=0 && newx=0 && newy

4.路径优化并查集

func find(_ x:Int,_ parent:[Int]) -> Int {
        var x1 = x
        while parent[x1] != x1 {
            x1 = parent[x1]
        }
        return x1
    }
    
    func union(_ i:Int,_ j:Int,_ parent:inout [Int],_ count:inout Int) {
        let x1 = find(i, parent)
        let y1 = find(j, parent)
        if x1 != y1 {//合并
            parent[x1] = y1
            count -= 1
        }
    }
    /*
     1. 访问过的状态置为0
     2. 在区间范围内，且是1可以连通
     */
    func numIslands(_ grid: [[Character]]) -> Int {
        let row = grid.count
        let col = grid[0].count
        
        var count = 0//记录1的个数
        var parent:[Int] = Array.init(repeating: 0, count: row*col)

        for i in 0..=0 && newX=0 && newY