305.Number of Islands II
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/* * 305.Number of Islands II * 2016-4-3 by Mingyang *Union Find 的题目,直接略过 */ private int[][] dir = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}}; public List<Integer> numIslands2(int m, int n, int[][] positions) { UnionFind2D islands = new UnionFind2D(m, n); List<Integer> ans = new ArrayList<Integer>(); for (int[] position : positions) { int x = position[0], y = position[1]; int p = islands.add(x, y); for (int[] d : dir) { int q = islands.getID(x + d[0], y + d[1]); if (q > 0 && !islands.find(p, q)) islands.unite(p, q); } ans.add(islands.size()); } return ans; } } class UnionFind2D { private int[] id; private int[] sz; private int m, n, count; public UnionFind2D(int m, int n) { this.count = 0; this.n = n; this.m = m; this.id = new int[m * n + 1]; this.sz = new int[m * n + 1]; } public int index(int x, int y) { return x * n + y + 1; } public int size() { return this.count; } public int getID(int x, int y) { if (0 <= x && x < m && 0<= y && y < n) return id[index(x, y)]; return 0; } public int add(int x, int y) { int i = index(x, y); id[i] = i; sz[i] = 1; ++count; return i; } public boolean find(int p, int q) { return root(p) == root(q); } public void unite(int p, int q) { int i = root(p), j = root(q); if (sz[i] < sz[j]) { //weighted quick union id[i] = j; sz[j] += sz[i]; } else { id[j] = i; sz[i] += sz[j]; } --count; } private int root(int i) { for (;i != id[i]; i = id[i]) id[i] = id[id[i]]; //path compression return i; } }
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leetcode305- Number of Islands II- hard