经典算法总结（持续完善中……）

基本参考自leetcode官方写法，希望能写出简洁、高效、标准的模式

二分查找 

    public int search(int[] nums, int target) {
        int low = 0, high = nums.length - 1;
        while (low <= high) {
            int mid = (high - low) / 2 + low;
            int num = nums[mid];
            if (num == target) {
                return mid;
            } else if (num > target) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return -1;
    }

参考：. - 力扣（LeetCode）

双指针

回文串判断： 

    //while形式
    public boolean isPalindrome(String s) {
        int left = 0, right = s.length() - 1;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            ++left;
            --right;
        }
        return true;
    }

    //for形式
    public boolean isPalindrome(String s) {
        for(int i =0, j=s.length()-1; i < j; i++, j--){
            if(s.charAt(i) != s.charAt(j)){
                return false;
            }
        }
        return true;
    }

参考：125. 验证回文串

二叉树：深度优先搜索

    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        } else {
            int leftHeight = maxDepth(root.left);
            int rightHeight = maxDepth(root.right);
            return Math.max(leftHeight, rightHeight) + 1;
        }
    }

参考：    104. 二叉树的最大深度：. - 力扣（LeetCode）

二叉树：广度优先搜索

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        if (root == null) {
            return ret;
        }

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> level = new ArrayList<Integer>();
            int currentLevelSize = queue.size();
            for (int i = 1; i <= currentLevelSize; ++i) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            ret.add(level);
        }
        
        return ret;
    }

 参考：102. 二叉树的层序遍历 ：. - 力扣（LeetCode）

图的深度优先搜索 

    void dfs(char[][] grid, int r, int c) {
        int nr = grid.length;
        int nc = grid[0].length;

        if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
            return;
        }

        grid[r][c] = '0';
        dfs(grid, r - 1, c);
        dfs(grid, r + 1, c);
        dfs(grid, r, c - 1);
        dfs(grid, r, c + 1);
    }

    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int nr = grid.length;
        int nc = grid[0].length;
        int num_islands = 0;
        for (int r = 0; r < nr; ++r) {
            for (int c = 0; c < nc; ++c) {
                if (grid[r][c] == '1') {
                    ++num_islands;
                    dfs(grid, r, c);
                }
            }
        }

        return num_islands;
    }

 参考： 200. 岛屿数量 ： . - 力扣（LeetCode）

图的广度优先搜索 

    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int nr = grid.length;
        int nc = grid[0].length;
        int num_islands = 0;

        for (int r = 0; r < nr; ++r) {
            for (int c = 0; c < nc; ++c) {
                if (grid[r][c] == '1') {
                    ++num_islands;
                    grid[r][c] = '0';
                    Queue<Integer> neighbors = new LinkedList<>();
                    neighbors.add(r * nc + c);
                    while (!neighbors.isEmpty()) {
                        int id = neighbors.remove();
                        int row = id / nc;
                        int col = id % nc;
                        if (row - 1 >= 0 && grid[row-1][col] == '1') {
                            neighbors.add((row-1) * nc + col);
                            grid[row-1][col] = '0';
                        }
                        if (row + 1 < nr && grid[row+1][col] == '1') {
                            neighbors.add((row+1) * nc + col);
                            grid[row+1][col] = '0';
                        }
                        if (col - 1 >= 0 && grid[row][col-1] == '1') {
                            neighbors.add(row * nc + col-1);
                            grid[row][col-1] = '0';
                        }
                        if (col + 1 < nc && grid[row][col+1] == '1') {
                            neighbors.add(row * nc + col+1);
                            grid[row][col+1] = '0';
                        }
                    }
                }
            }
        }

        return num_islands;
    }

 参考：    200. 岛屿数量 ： . - 力扣（LeetCode）