下载源代码后，解压。 下载lcov 安装lcov make install。 完成后，运行./buildAndRun.sh即可，最后运行测试用例，生成覆盖率报告。
如果不想安装lcov或者不想生成覆盖率报告，则可以将 buidlAndRun.sh中的以下两句，前面加#注释掉
#lcov -c -o main.info -d .
#生成html
#genhtml main.info -o ./gcov
the answers to the leetcode questions
├── gtest //google test
├── leetcode //source code
├── gcov //coverage report
├── main //run file
├── Makefile // Make file
├── README.md //
├── buildAndRun.sh //run shell
└── testcase //test case file
g++ -o 001_two_sum.o -c ./leetcode/src/001_Two_Sum.c -I ./leetcode/include/
g++ -o leetcode_test.o -c ./testcase/src/leetcode_test.cpp -I ./ -I ./leetcode/include/
g++ -o TestAll.o -c ./gtest/main/TestAll.cpp -I ./
g++ -o main *.o -I./include -L./gtest/lib -lgtest -lpthread\
to run the test ,you can execute the commond
./buildAndRun.sh
the article link ：http://www.huyanbing.me/2017/08/04/leetcode%E9%A2%98%E7%9B%AE%E8%A7%A3%E6%9E%90/
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example: Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
给定整型数组，返回两个数的下标，使得这两个数相加得到特定的值。 假设每个给定的数组只能找到一组满足条件的结果，同时，不能使用同一个数两次。
题大意为，在一组数组中，找到两个数，使得这两个数的和等于特定值，并返回下标。可以从第一个数开始，循环与后面的每一个相加，与结果比较，比较成功则返回。 例如，输入[1,7,11,15],目标值26，那么循环计算1+7,1+11,1+15，7+11,7+15......，直到得到目标值。
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8
给定两个非空链表代表两个非负整数，整数的各位数以逆序存储在链表的每个节点中。将这两个数相加，并返回结果链表。
题意较清晰，是将用链表形式的两个整数进行相加，并返回链表结果。 需要注意的主要有以下几点 1.加完之后需要给下一位子进位。 2.如果链表只有一位，直接计算结果，提高效率。 3.考虑两个链表长度不一样的场景
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
计算各位数之和，直到最后只有一位。
计算各位数之和，采用递归，计算一次后，再次调用，直到结果位各位数。
Write a function that takes a string as input and returns the string reversed.
Example: Given s = "hello", return "olleh".
字符串翻转
头尾对应位置的字符位置调换
The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
Given two integers x and y, calculate the Hamming distance.
Note: 0 ≤ x, y < 231.
Example:
Input: x = 1, y = 4
Output: 2
Explanation: 1 (0 0 0 1) 4 (0 1 0 0) ? ?
The above arrows point to positions where the corresponding bits are different.
两个数的汉明距离，可以理解为，二进制的情况下，两个数异或之后的数的1的个数。 比如例子中，1和4,0001与0100异或得：0101，而0101中1的个数，即为汉明距离，可以理解位，从0001，变成0100，需要改变的位数。
Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
Note: The given integer is guaranteed to fit within the range of a 32-bit signed integer. You could assume no leading zero bit in the integer’s binary representation. Input: 5 Output: 2 Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
给定一个正整数，输出它的补全整数。补全策略是翻转它的每个比特位，得到补全整数。
对于补全，以5为例，101，每一位翻转得到，010，即结果为2，那么，010+101 = 111，2的3次方。再以35为例，10 0011，翻转得到011100，即有100011+011100=111111,2的6次方，那么不难得到，其实补全数，就是用2的n次方，减去该数，其中，n为该数二进制表示的位数。