《剑指Offer》2.3数据结构 java版

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No.3 二维数组中查找数字
思路：从右上角（左下角）开始比较

    public boolean Find(int target, int [][] array) {        if(array == null)            return false;        int row = array.length, column = array[0].length;        int i = 0, j = column - 1;        while( i < row && j >= 0){            if(array[i][j] == target)                return true;            if(array[i][j] > target)                j--;            else i++;        }        return false;    }

No.4：将字符串中的空格替换为”%20“
思路：第一种：可以新开辟空间存储；第二种，在原字符串上修改，从后往前

    public String replaceSpace(StringBuffer str) {        if(str == null || str.length() == 0)            return str.toString();        int len = str.length();        StringBuilder sb = new StringBuilder();        for(int i = 0; i < len; i++){            if(str.charAt(i) == ' '){                sb.append("%20");            }else{                sb.append(str.charAt(i));            }        }        return sb.toString();    }

推广：将两个数组按序合并，在原位置

    public int[] mergerDigital(int[] A,int[] B, int n1, int n2){        if(n2 == 0)            return A;        if(n1 == 0){            A = Arrays.copyOf(B, n2);            return A;        }        int k = n1 + n2 - 1, i = n1 - 1, j = n2 - 1;        while(i >= 0 && j >= 0){            A[k--] = A[i] >= B[j] ? A[i--] : B[j--];        }        while(i >= 0){            A[k--] = A[i--];        }        while(j >= 0){            A[k--] = B[j--];        }        return A;    }

No.5 将单向链表逆序输出
思路：使用栈，后进先出；或者使用linkedlist

    public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {        Stack s = new Stack();        while(listNode != null){            s.push(listNode.val);            listNode = listNode.next;        }        ArrayList al = new ArrayList();        while(!s.empty()){            al.add(s.pop());        }        return al;    }

    public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {        LinkedList s = new LinkedList();        while(listNode != null){            s.addFirst(listNode.val);            listNode = listNode.next;        }        return new ArrayList<>(s);    }

No.6 重建二叉树
思路：递归思想

    public TreeNode help(int[] preorder, int[] inorder, int pleft, int pright, int ileft, int iright) {        if(pleft > pright || ileft > iright)            return null;        TreeNode root = new TreeNode(preorder[pleft]);        int i;        for(i = ileft; i <= iright; i++){            if(inorder[i] == preorder[pleft])                break;        }        root.left = help(preorder, inorder,pleft + 1, pleft + i - ileft, ileft, i - 1);        root.right = help(preorder, inorder,pleft + i - ileft + 1, pright, i + 1, iright);        return root;    }    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {        int len = pre.length;        if(len == 0)            return null;        return help(pre, in, 0, len - 1, 0, len - 1);    }

No.7 用两个栈实现一个队列

        Stack<Integer> stack1 = new Stack<Integer>();        Stack<Integer> stack2 = new Stack<Integer>();        public void push(int node) {            stack1.push(node);        }        public int pop() throws Exception {            if(stack2.isEmpty()){                while(!stack1.isEmpty()){                    stack2.push(stack1.pop());                }            }            if(stack2.isEmpty())                throw new Exception("队列为空，不能删除");            return stack2.pop();        }

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