劍指offer 1

1 二維數(shù)組中的查找

public class Solution {
    public boolean Find(int target, int [][] array) {
        if (array.length == 0) {
            return false;
        }
        int len1 = array.length;//求行
        int len2 = array[0].length;//求列
        for (int j = len2 - 1; j >= 0; j--) {
            for (int i = 0; i <= len1 - 1; i++) {
                if (target < array[i][j]) {
                    break;
                }else if (target > array[i][j]) {
                    continue;
                }else {
                    return true;
                }
            }
        }
        return false;
    }
}

2 替換字符串

public class Solution {
    public String replaceSpace(StringBuffer str) {
        if (str == null) {
            return null;
        }
        /*
        if (str.length() == 0) {//出錯(cuò)
            return null;
        }
        */
        for (int i = 0; i < str.length(); i++) {
            char c = str.charAt(i);
            if (c == ' ') {
                str.replace(i,i+1,"%20");
            }
        }
        return str.toString();
    }
}

3 從尾到頭打印鏈表

import java.util.Stack;//注意添加包
import java.util.ArrayList;
public class Solution {
    public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
        if (listNode == null) {
            ArrayList<Integer> list = new ArrayList<>();
            return list;
        }
        Stack<Integer> stack = new Stack<>();
        while (listNode != null) {
            stack.push(listNode.val);
            listNode = listNode.next;
        }
        ArrayList<Integer> list = new ArrayList<>();
        while (!stack.isEmpty()) {
            list.add(stack.pop());
        }
        return list;
    }
}

4 重建二叉樹

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
         if(pre.length == 0||in.length == 0){
            return null;
        }
        TreeNode node = new TreeNode(pre[0]);
        for(int i = 0; i < in.length; i++){
            if(pre[0] == in[i]){
                node.left = reConstructBinaryTree(Arrays.copyOfRange(pre, 1, i+1), Arrays.copyOfRange(in, 0, i));
                node.right = reConstructBinaryTree(Arrays.copyOfRange(pre, i+1, pre.length), Arrays.copyOfRange(in, i+1,in.length));
            }
        }
        return node;
    }
}

5 兩個(gè)棧實(shí)現(xiàn)隊(duì)列

import java.util.Stack;

public class Solution {
    Stack<Integer> stack1 = new Stack<Integer>();
    Stack<Integer> stack2 = new Stack<Integer>();
    
    public void push(int node) {
        stack1.push(node);
    }
    
    public int pop() {
        if (stack2.isEmpty()) {//只有當(dāng)stack2為空時(shí)才一次性將stack1中的數(shù)據(jù)全部壓入
            while (!stack1.isEmpty()) {
                stack2.push(stack1.pop());
            }
        }
        return stack2.pop();
    }
}

6 旋轉(zhuǎn)數(shù)組的最小數(shù)字

//二分查找
import java.util.ArrayList;
public class Solution {
    public int minNumberInRotateArray(int [] array) {
        if (array.length == 0) {
            return 0;
        }
        int low = 0;
        int high = array.length - 1;
        while (low < high) {
            int mid = (low + high) / 2;
            if (array[mid] > array[high]) {//注意是拿mid和high比
                low = mid + 1;
            }else if (array[mid] < array[high]) {
                high = mid;
            }else {
                high -= 1;
            }
        }
        return array[low];
    }
}

7 斐波那契數(shù)列

//1,1,2,3,5,8,13,21
public class Solution {
    public int Fibonacci(int n) {
        if (n == 0) {
            return 0;
        }
        int a = 1, b = 1, c = 0;
        if (n == 1 || n == 2) {
            return 1;
        }else {
            for (int i = 3; i <= n; i++) {
                c = a + b;
                a = b;
                b = c;
            }
        }
        return c;
    }
}

8 跳臺(tái)階

//變版斐波那契數(shù)列:1,2,3,5,8,13,21
//當(dāng)前臺(tái)階的跳法總數(shù)=當(dāng)前臺(tái)階后退一階的臺(tái)階的跳法總數(shù)+當(dāng)前臺(tái)階后退二階的臺(tái)階的跳法總數(shù)
//n級(jí)=n-1級(jí)的每個(gè)方法后面添上1,n-2級(jí)的每個(gè)方法后面填上2

public class Solution {
    public int JumpFloor(int target) {
        if (target <= 0) {
            return 0;
        }
        if (target == 1) {
            return 1;
        }
        if (target == 2) {
            return 2;
        }
        int a = 1;
        int b = 2;
        int c = 0;
        for (int i = 3; i <= target; i++) {
            c = a + b;
            a = b;
            b = c;
        }
        return c;
    }
}

9 變態(tài)跳臺(tái)階

//n = 之前所以方法的和再加1,
//1,2,4,8,16,32
public class Solution {
    public int JumpFloorII(int target) {
        if (target <= 0) {
            return 0;
        }
        if (target == 1) {
            return 1;
        }
        if (target == 2) {
            return 2;
        }
        int sum = 3;
        int c = 0;
        for (int i = 3; i <= target; i++) {
            sum = sum + c;
            c = sum + 1;
        }
        return c;
    }
}

10 矩形覆蓋

//變斐波那契:1,2,3,5,8,13,21;同跳臺(tái)階
public class Solution {
    public int RectCover(int target) {
        if (target <= 0) {
            return 0;
        }
        if (target == 1) {
            return 1;
        }
        if (target == 2) {
            return 2;
        }
        int a = 1;
        int b = 2;
        int c = 0;
        for (int i = 3; i <= target; i++) {
            c = a + b;
            a = b; 
            b = c;
        }
        return c;
    }
}
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