摘要：面試題從尾到頭打印鏈表輸入一個鏈表，從尾到頭打印鏈表每個節(jié)點的值面試題重建二叉樹輸入某二叉樹的前序遍歷和中序遍歷的結(jié)果，請重建出該二叉樹。例如輸入前序遍歷序列和中序遍歷序列，則重建二叉樹并返回。隊列中的元素為類型。其中負數(shù)用補碼表示。

public boolean find(int target, int [][] array) {
    boolean found = false;
    
    if(array == null || array.length == 0){
        return false;
    }
    
    int rows = array.length;
    int columns = array[0].length;
    
    if(rows > 0 && columns > 0){
        int row = 0;
        int column = columns - 1;
        while(row < rows && column >= 0){
            if(array[row][column] == target){
                found = true;
                break;
            }else if(array[row][column] > target){
                column--;
            }else{
                row++;
            }
        } 
    }
    return found;
}

請實現(xiàn)一個函數(shù)，將一個字符串中的空格替換成“%20”。例如，當字符串為We Are Happy.則經(jīng)過替換之后的字符串為We%20Are%20Happy。

public String replaceSpace(StringBuffer str) {
    if(str == null){
        return null;
    }
    
    int spaceCount = 0;
    int length = str.length();
    for(int i = 0; i < length; i++){
        if(str.charAt(i) == " "){
            spaceCount++;
        }
    }
    int newLength = length + 2 * spaceCount;
    char[] newStr = new char[newLength];
    for(int i = length - 1; i >= 0; i--){
        if(str.charAt(i) == " "){
            newStr[--newLength] = "0";
            newStr[--newLength] = "2";
            newStr[--newLength] = "%";
        }else{
            newStr[--newLength] = str.charAt(i);
        }
    }
    return new String(newStr);
}

輸入一個鏈表，從尾到頭打印鏈表每個節(jié)點的值

    public ArrayList printListFromTailToHead(ListNode listNode) {
        if(listNode == null){
            return new ArrayList();
        }
        Stack stack = new Stack<>(); 
        ArrayList res = new ArrayList<>();
        while(listNode != null){
            stack.push(listNode);
            listNode = listNode.next;
        }
        
        while(!stack.isEmpty()){
            ListNode cur = stack.pop();
            res.add(cur.val);
        }
        return res;
    }

輸入某二叉樹的前序遍歷和中序遍歷的結(jié)果，請重建出該二叉樹。假設輸入的前序遍歷和中序遍歷的結(jié)果中都不含重復的數(shù)字。例如輸入前序遍歷序列{1,2,4,7,3,5,6,8}和中序遍歷序列{4,7,2,1,5,3,8,6}，則重建二叉樹并返回。

    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
        if(pre.length == 0 || in.length == 0){
            return null;
        }
        
        TreeNode root = new TreeNode(pre[0]);
        int len = pre.length;
        int inorderPos = 0;
        for(;inorderPos < in.length; inorderPos++){
            if(in[inorderPos] == root.val){
                break;
            }
        }
        
        int leftTreeLength = inorderPos;
        int[] leftInorder = new int[leftTreeLength];
        int[] leftPre = new int[leftTreeLength];
        for(int i = 0; i < leftTreeLength; i++){
            leftInorder[i] = in[i];
            leftPre[i] = pre[i + 1];
        }
        
        int rightTreeLength = len - leftTreeLength - 1;
        int[] rightInorder = new int[rightTreeLength];
        int[] rightPre = new int[rightTreeLength];
        
        for(int i = 0; i < rightTreeLength; i++){
            rightInorder[i] = in[inorderPos + 1 + i];
            rightPre[i] = pre[leftTreeLength + 1 + i];
        }
        
        TreeNode left = reConstructBinaryTree(leftPre, leftInorder);
        TreeNode right = reConstructBinaryTree(rightPre, rightInorder);
        root.left = left;
        root.right = right;
        
        return root;
        
    }

用兩個棧來實現(xiàn)一個隊列，完成隊列的Push和Pop操作。 隊列中的元素為int類型。

    Stack stack1 = new Stack();
    Stack stack2 = new Stack();
    
    public void push(int node) {
        stack1.push(node);
    }
    
    public int pop() {
        if(stack2.isEmpty()){
            while(!stack1.isEmpty()){
                int cur = stack1.pop();
                stack2.push(cur);
            }
        }
        return stack2.pop();
    }

把一個數(shù)組最開始的若干個元素搬到數(shù)組的末尾，我們稱之為數(shù)組的旋轉(zhuǎn)。 輸入一個非遞減排序的數(shù)組的一個旋轉(zhuǎn)，輸出旋轉(zhuǎn)數(shù)組的最小元素。 例如數(shù)組{3,4,5,1,2}為{1,2,3,4,5}的一個旋轉(zhuǎn)，該數(shù)組的最小值為1。 NOTE：給出的所有元素都大于0，若數(shù)組大小為0，請返回0。

 public int minNumberInRotateArray(int [] array) {
        if(array == null || array.length == 0){
            return 0;
        }
        int length = array.length;
        int index1 = 0;
        int index2 = length - 1;
        int indexMid = index1;
        while(array[index1] >= array[index2]){
            if(index2 - index1 == 1){
                indexMid = index2;
                break;
            }
            System.out.println(index1 + " ***** " + index2);
            indexMid = (index1 + index2) / 2;
            if(array[index1] == array[index2] && array[indexMid] == array[index1]){
                return minInOrder(array, index1, index2);
            }
            if(array[indexMid] >= array[index1]){
                index1 = indexMid;
            }else if(array[indexMid] <= array[index2]){
                index2 = indexMid;
            }
        }
        return array[indexMid];
    }
    
    public int minInOrder(int[] array, int index1, int index2){
        int result = array[index1];
        for(int i = index1 + 1; i <= index2; i++){
            if(result > array[i]){
                result = array[i];
            }
        }
        return result;
    }

大家都知道斐波那契數(shù)列，現(xiàn)在要求輸入一個整數(shù)n，請你輸出斐波那契數(shù)列的第n項。
n<=39

    public int fibonacci(int n) {
        int[] res = new int[]{0, 1};
        if(n < 2){
            return res[n];
        }
        
        int fibN = 0;
        int fibNMinusOne = 1;
        int fibNMinusTwo = 0;
        for(int i = 2; i <= n; i++){
            fibN = fibNMinusOne + fibNMinusTwo;
            fibNMinusTwo = fibNMinusOne;
            fibNMinusOne = fibN;
        }
        return fibN;
    }

輸入一個整數(shù)，輸出該數(shù)二進制表示中1的個數(shù)。其中負數(shù)用補碼表示。

    public int numberOf1(int n) {
        int count = 0;
        int flag = 1;
        while(flag != 0){
            if((n & flag) != 0){
                count++;
            }
            flag = flag << 1;
        }
        return count;
    }