力扣---2020.3.22

945. 使数组唯一的最小增量

class Solution {
    public int minIncrementForUnique(int[] A) {
        Arrays.sort(A);
        int move = 0;
        for(int i =1;i<A.length;i++){
            if(A[i]<=A[i-1]){
                int temp = A[i];
                A[i] = A[i-1] + 1;
                move += A[i] - temp;
            }
        }
        return move;
    }
}

//线性探测法+路径压缩O(N)
class Solution {
    int[] pos = new int [80000];
    public int minIncrementForUnique(int[] A) {
        Arrays.fill(pos, -1); // -1表示空位
        int move = 0;
        // 遍历每个数字a对其寻地址得到位置b, b比a的增量就是操作数。
        for (int a: A) {
            int b = findPos(a); 
            move += b - a;
        }
        return move;
    }
    
    // 线性探测寻址（含路径压缩）
    private int findPos(int a) {
        int b = pos[a];
        // 如果a对应的位置pos[a]是空位，直接放入即可。
        if (b == -1) { 
            pos[a] = a;
            return a;
        }
        // 否则向后寻址
        // 因为pos[a]中标记了上次寻址得到的空位，因此从pos[a]+1开始寻址就行了（不需要从a+1开始）。
        b = findPos(b + 1); 
        pos[a] = b; // 寻址后的新空位要重新赋值给pos[a]哦，路径压缩就是体现在这里。
        return b;
    }
}

//计数
class Solution {
    public int minIncrementForUnique(int[] A) {
        int[] count = new int[80000];
        for (int x: A) count[x]++;

        int ans = 0, taken = 0;

        for (int x = 0; x < 80000; ++x) {
            if (count[x] >= 2) {
                taken += count[x] - 1;
                ans -= x * (count[x] - 1);
            }
            else if (taken > 0 && count[x] == 0) {
                taken--;
                ans += x;
            }
        }

        return ans;
    }
}

面试题68 - I. 二叉搜索树的最近公共祖先

//递归
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null) return null;
        if(root.val>p.val && root.val>q.val){
            return lowestCommonAncestor(root.left,p,q);
        }
        if(root.val<p.val && root.val<q.val){
            return lowestCommonAncestor(root.right,p,q);
        }
        return root;
    }
}

//非递归
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while (root!=null){
            if (p.val<root.val && q.val<root.val){
                root=root.left;
            }else if (p.val>root.val && q.val>root.val){
                root=root.right;
            }else{
                break;
            }
        }
        return root;
    }
}

面试题07. 重建二叉树

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int n = preorder.length;
        if(n == 0) return null;
        int rootVal = preorder[0],rootIndex = 0;
        for(int i=0;i<n;i++){
            if(inorder[i] == rootVal){
                rootIndex = i;
                break;
            }
        }
        TreeNode root = new TreeNode(rootVal);
        root.left = buildTree(Arrays.copyOfRange(preorder, 1, 1 + rootIndex), Arrays.copyOfRange(inorder, 0, rootIndex));
        root.right = buildTree(Arrays.copyOfRange(preorder, 1 + rootIndex, n), Arrays.copyOfRange(inorder, rootIndex + 1, n));
        return root;
    }
}

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        Map<Integer, Integer> indexMap = new HashMap<Integer, Integer>();
        int length = preorder.length;
        for (int i = 0; i < length; i++) {
            indexMap.put(inorder[i], i);
        }
        TreeNode root = buildTree(preorder, 0, length - 1, inorder, 0, length - 1, indexMap);
        return root;
    }

    public TreeNode buildTree(int[] preorder, int preorderStart, int preorderEnd, int[] inorder, int inorderStart, int inorderEnd, Map<Integer, Integer> indexMap) {
        if (preorderStart > preorderEnd) {
            return null;
        }
        int rootVal = preorder[preorderStart];
        TreeNode root = new TreeNode(rootVal);
        if (preorderStart == preorderEnd) {
            return root;
        } else {
            int rootIndex = indexMap.get(rootVal);
            int leftNodes = rootIndex - inorderStart, rightNodes = inorderEnd - rootIndex;
            TreeNode leftSubtree = buildTree(preorder, preorderStart + 1, preorderStart + leftNodes, inorder, inorderStart, rootIndex - 1, indexMap);
            TreeNode rightSubtree = buildTree(preorder, preorderEnd - rightNodes + 1, preorderEnd, inorder, rootIndex + 1, inorderEnd, indexMap);
            root.left = leftSubtree;
            root.right = rightSubtree;
            return root;
        }
    }
}

//迭代
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[0]);
        int length = preorder.length;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        int inorderIndex = 0;
        for (int i = 1; i < length; i++) {
            int preorderVal = preorder[i];
            TreeNode node = stack.peek();
            if (node.val != inorder[inorderIndex]) {
                node.left = new TreeNode(preorderVal);
                stack.push(node.left);
            } else {
                while (!stack.isEmpty() && stack.peek().val == inorder[inorderIndex]) {
                    node = stack.pop();
                    inorderIndex++;
                }
                node.right = new TreeNode(preorderVal);
                stack.push(node.right);
            }
        }
        return root;
    }
}

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