benbenab的专栏

Implement int sqrt(int x).Compute and return the square root of x.二分法：class Solution {public: int sqrt(int x) { // Start typing your C/C++ solution below //

class Solution {public: vector grayCode(int n) { vector results; do{ if (n < 0) break; else if (n == 0) { resul

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.For example, given the array [−2,1,−3,4,−1,2,1,−5,4],the contiguous subarray [4,−1,2,1] ha

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.You may assume no duplicates in the array.

Reverse digits of an integer.Example1: x = 123, return 321Example2: x = -123, return -321Have you thought about this?Here are some good questions to ask before coding. Bonus points for

class Solution {public: int divide(int dividend, int divisor) { // Start typing your C/C++ solution below // DO NOT write int main() function if (divisor == 0)

Given a sorted array of integers, find the starting and ending position of a given target value.Your algorithm's runtime complexity must be in the order of O(log n).If the target is not found in t

不知道这个的复杂度怎么分析。 不过worst case 应该是O（n），average 可能是O(sqrt(n))int maxPrime(int n){ assert(n>=0); if(n <= 3) return n; int k = 1; while(n > 1) { ++k; if(k > n / k) ret

1. 考虑double float 的相等，不能仅仅用==表示，是有精度限制的。2.  n 大于0，小于0的情况3. if (temp_diveded & 1) result *= base; base *= base; temp_diveded >>= 1;想法很赞！！！！#define AC

1. integer到roman利用了数组的位置的信息class Solution {public: string intToRoman(int num) { // Start typing your C/C++ solution below // DO NOT write int main() function char

Answers write code in java /c for expression 5+4*(7-15) or have parenthesis in any order .idea: Use two stack, one for number, one for operator. Every time push a number or an operation into stack.

A message containing letters from A-Z is being encoded to numbers using the following mapping:'A' -> 1'B' -> 2...'Z' -> 26Given an encoded message containing digits, determine the total nu

方法讨论：http://blog.kingsamchen.com/archives/649typedef struct{ int value; int index;}info;bool less_b (const info& info1, const info& info2){ return info1.value < info2.value;}int * temp = new

4SumGiven an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.No

数组中有一个数字出现了超过数组长度的一般，请找出这个数字。思路：如果这个数子长度超过一半，那肯定排序后他会出现在middle的位置上！int Partition(int* num, unsigned int length, int start, int end){ int pivot = num[end]; int i = start-1; int index; for

1.  数值的整数次方   a. 不考虑大数情况 #define PRECISION 0.00001double pow(double base, int exp){ if (equal(base,0)) return 0; if (exp == 0) return 1.0; bool negative = (exp>0)?false:true;

Single NumberGiven an array of integers, every element appears twice except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it