1. 求一个数组中后边的元素减去前边的元素的最大值
例如
数组 [3, 2, -1, 5, 4]
后减前最大值为:5 - (-1)=6
思路
定义局部最大值tmpMax(初始化为负无穷),局部最小值tmpLow(初始化为数组第一个元素)
从第二个元素开始遍历,如果(该元素-tmpLow )< 0说明该元素比tmpLow还小呢,那么就把tmpLow替换成该元素,同时比较该差值与tmpMax的大小,用大的替换tmpMax;否则(>=0)只需比较该差值与tmpMax的大小,用大的替换tmpMax。
参考代码
#include <iostream>
using namespace std;
int MAX_INT = ((unsigned)(-1)) >> 1;
int MIN_INT = -MAX_INT;
int getMaxGap(int A[], int lens)
{
if (A == NULL || lens <= 0)
return -1;
int tmpMax = MIN_INT;
int tmpLow = A[0];
for (int i = 1; i < lens; ++i)
{
int tmp = A[i] - tmpLow;
if (tmp > tmpMax)
tmpMax = tmp;
if (tmp < 0)
tmpLow = A[i];
}
return tmpMax;
}
int main()
{
int a[] = {3, 2, -1, 5, 4};
cout << getMaxGap(a, sizeof(a) / sizeof(int)) << endl;
}
结果
6
2. 删除字符串开始及末尾的空白符,并且把数组中间的多个空格(如果有)符转化为1个。
思路:两个指针往后走
参考代码
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;
void getNewStr(char a[])
{
int new_cur = 0;
int Isbeg = 1;
int i = 0;
for (i = 0; a[i] != '\0'; )
{
if (a[i] != ' ')
{
Isbeg = 0;
a[new_cur++] = a[i++];
}
else
{
if (Isbeg == 1)
{
while (a[i] != '\0' && a[i] == ' ')
++i;
}
else
{
a[new_cur++] = a[i++];
while (a[i] != '\0' && a[i] == ' ')
++i;
}
}
}
cout << "new_cur:" << new_cur << endl;
if (a[i-1] == ' ')
a[new_cur-1] = '\0';
else
a[new_cur] = '\0';
}
int main()
{
char a[] = " hello hi ";
cout << a << endl;
cout << sizeof(a) / sizeof(char) << endl;
getNewStr(a);
cout << a << endl;
cout << sizeof(a) / sizeof(char) << endl;
for (int i = 0; a[i] != '\0'; ++i)
cout << a[i] << endl;
}