牛客网剑指offer

Author: JavaScalaDeveloper

jianzhioffer/src/jz_offer/.gitignore

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+### IDEA ###
+.idea/
+*.iml
+target/
+*/target/
+*/*.iml
+*/.gradle/
+*/out/
+out/
+### Java ###
+.settings/
+.classpath
+.project
+bin/
+build/

jianzhioffer/src/jz_offer/LICENSE

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+MIT License
+
+Copyright (c) 2020 David
+
+Permission is hereby granted, free of charge, to any person obtaining a copy
+of this software and associated documentation files (the "Software"), to deal
+in the Software without restriction, including without limitation the rights
+to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
+copies of the Software, and to permit persons to whom the Software is
+furnished to do so, subject to the following conditions:
+
+The above copyright notice and this permission notice shall be included in all
+copies or substantial portions of the Software.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
+FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
+AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
+LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
+OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
+SOFTWARE.

jianzhioffer/src/jz_offer/README.md

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+# 牛客网剑指Offer题解
+
+#### 介绍
+牛客网剑指Offer题解(java实现)
+

jianzhioffer/src/jz_offer/all/JZ1/Solution.java

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+package jz_offer.all.JZ1;
+
+public class Solution {
+    //暴力法
+    /*public boolean Find(int target, int [][] array) {
+        if(array.length==0)
+            return false;
+        for(int i=0;i<array.length;i++){
+            for(int j=0;j<array[0].length;j++){
+                if(array[i][j]==target)
+                    return true;
+            }
+        }
+        return false;
+    }*/
+
+    //时间复杂度相比暴力法较低
+    public boolean Find(int target, int[][] array) {
+        int row = array.length;
+        if (row == 0)
+            return false;
+        int col = array[0].length;
+        if (col == 0)
+            return false;
+        int rows = row - 1;
+        int cols = 0;
+        while (rows >= 0 && cols < col) {
+            if (target < array[rows][cols])
+                rows--;
+            else if (target > array[rows][cols])
+                cols++;
+            else
+                return true;
+        }
+        return false;
+    }
+}

jianzhioffer/src/jz_offer/all/JZ1/二维数组中的查找

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+题目描述：
+在一个二维数组中（每个一维数组的长度相同），每一行都按照从左到右递增的顺序排序，
+每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，
+判断数组中是否含有该整数。
+
+输入：7,[[1,2,8,9],[2,4,9,12],[4,7,10,13],[6,8,11,15]]   返回：true
+
+思路：
+1、时间复杂度：算法中基本操作的执行次数
+2、一种暴力求解
+3、另一种由于行从左到右递增，列从上到下递增，可以用这个性质。这样时间复杂度就变成行+列的次数，
+时间复杂度大大减小

jianzhioffer/src/jz_offer/all/JZ10/Solution.java

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+package jz_offer.all.JZ10;
+
+public class Solution {
+    public int RectCover(int target) {
+        if (target == 1 || target == 2)
+            return target;
+        int one = 1;
+        int two = 2;
+        int result = 0;
+        for (int i = 2; i < target; i++) {
+            result = one + two;
+            one = two;
+            two = result;
+        }
+        return result;
+    }
+}

jianzhioffer/src/jz_offer/all/JZ10/矩形覆盖

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+题目描述：
+我们可以用2*1的小矩形横着或者竖着去覆盖更大的矩形。
+请问用n个2*1的小矩形无重叠地覆盖一个2*n的大矩形，总共有多少种方法？
+
+比如n=3时，2*3的矩形块有3种覆盖方法：
+
+思路：
+1、找规律，可以用递归或者递推，一般使用递推。因为递归的时间复杂度比较高O(2^n)
+2、写出n=1到5时的情况，就可以找出规律

jianzhioffer/src/jz_offer/all/JZ11/Solution.java

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+package jz_offer.all.JZ11;
+
+public class Solution {
+    public int NumberOf1(int n) {
+        int count = 0;
+        if (n == 0)
+            return 0;
+        while (n != 0) {
+            if (n < 0)
+                ++count;
+            n = (n << 1);
+        }
+        return count;
+    }
+
+    //另一种方法，返回整数的二进制数，然后进行判断
+    //数值用==比较，字符串用equals比较，字符char用=='1'
+    /*public int NumberOf1(int n) {
+        String str=Integer.toBinaryString(n);
+        int count=0;
+        for(int i=0;i<str.length();i++){
+            if(str.charAt(i)=='1')
+                ++count;
+        }
+        return count;
+    }*/
+}

jianzhioffer/src/jz_offer/all/JZ11/二进制中1的个数

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+题目描述：输入一个整数，输出该数32位二进制表示中1的个数。其中负数用补码表示。
+
+思路：
+1、正数的原码、反码、补码是一样的，负数则有区别
+    计算机的整数计算用反码表示，因为反码可以表示加减，范围则是-128-127
+    采用反码计算，可以轻松得到答案
+    符号位为0，为正；符号位为1，为负
+    负数的反码就是原码取反，补码就是反码加1
+    原码首位为0是正数，首位为1是负数，因此这道题可以通过左移判断
+

jianzhioffer/src/jz_offer/all/JZ12/Solution.java

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+package jz_offer.all.JZ12;
+
+public class Solution {
+    public double Power1(double base, int exponent) {
+        return Math.pow(base, exponent);
+    }
+
+    public double Power2(double base, int exponent) {
+        double pow = 1.00;
+        if (base == 0 && exponent != 0)
+            return 0.00;
+        if (exponent == 0 && base != 0)
+            return 1.00;
+        if (exponent > 0) {
+            for (int i = 0; i < exponent; i++)
+                pow *= base;
+        } else if (exponent < 0) {
+            for (int i = 0; i < Math.abs(exponent); i++) {
+                pow /= base;
+            }
+        }
+        return pow;
+    }
+}