劍指offer之和為s的數(shù)組
1 問(wèn)題
輸入一個(gè)遞增排序數(shù)組和數(shù)字和s,在數(shù)組里面找2個(gè)數(shù),他們的和是s,如果有多對(duì),只需要輸出一對(duì)。
比如數(shù)組{1, 2, 4, 7, 11, 15},我們輸出4 ,11
2 思路
我們定義2個(gè)首尾指針,先是1+15,大于15,然后我們尾巴指針左移一下,然后就是1+11 小于15,然后首指針右移動(dòng)一下,2+11,依次類推。
3 代碼實(shí)現(xiàn)
#include <stdio.h>
#include <stdlib.h>
#define true 1
#define false 0
int findNumber(int *a, int len, int sum, int *num1, int *num2)
{
int result = false;
if (NULL == a || len < 1 || NULL == num1 || NULL == num2)
{
return result;
}
int start = 0;
int end = len - 1;
while (end > start)
{
*num1 = a[start];
*num2 = a[end];
int curSum = *num1 + *num2;
if (curSum == sum)
{
result = true;
break;
}
else if (curSum > sum)
{
--end;
}
else
{
++start;
}
}
return result;
}
int main()
{
int a[] = {1, 2, 4, 7, 11, 15};
int num1 = 1;
int num2 = 3;
int result = findNumber(a, sizeof(a) / sizeof(int), 15, &num1, &num2);
if (result == 1)
{
printf("the num1 is %d the num2 is %d\n", num1, num2);
}
else
{
printf("do not find the two number in numbers");
}
return 0;
}
4 運(yùn)行結(jié)果
the num1 is 4 the num2 is 11
5 代碼優(yōu)化
while (end > start)
{
int curSum = a[start] + a[end];
if (curSum == sum)
{
*num1 = a[start];
*num2 = a[end];
result = true;
break;
}
else if (curSum > sum)
{
--end;
}
else
{
++start;
}
}
作者:chen.yu
深信服三年半工作經(jīng)驗(yàn),目前就職游戲廠商,希望能和大家交流和學(xué)習(xí),
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