lintcode:Interleaving Positive and Negative Numbers

最新推荐文章于 2021-01-01 11:33:57 发布

martin_liang

最新推荐文章于 2021-01-01 11:33:57 发布

阅读量417

点赞数

CC 4.0 BY-SA版权

分类专栏：算法 C++/C lintcode

本文链接：https://blog.youkuaiyun.com/martin_liang/article/details/50859868

算法同时被 3 个专栏收录

537 篇文章

订阅专栏

C++/C

504 篇文章

订阅专栏

lintcode

66 篇文章

订阅专栏

本文介绍了一种算法，用于将包含正负整数的数组重新排列成正负数交错的形式，不需要保持原始顺序。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Interleaving Positive and Negative Numbers

You have exceeded the time limit

Reset

Given an array with positive and negative integers. Re-range it to interleaving with positive and negative integers.

Have you met this question in a real interview?

Yes

Notice

You are not necessary to keep the original order of positive integers or negative integers.

Given [-1, -2, -3, 4, 5, 6], after re-range, it will be [-1, 5, -2, 4, -3, 6]or any other reasonable answer.

这道题一定要做两次循环

第一次循环判断出正数多还是负数多，多的一个种数要放在［0］号位

class Solution {
public:
    /**
     * @param A: An integer array.
     * @return: void
     */
void rerange(vector<int> &A) {
   // write your code here
   int negativeIdx = 0;
   int positiveIdx = 1;
   int positiveNum = 0;
   
   if (A.size() % 2)
   {
        for (int i=0; i<A.size(); i++)
            if (A[i] > 0)
                positiveNum++;
                
        if (positiveNum > (A.size()/2))
        {
            positiveIdx = 0;
            negativeIdx = 1;
        }
   }
   
   while (negativeIdx < A.size() && positiveIdx < A.size())
   {
      while ((negativeIdx < A.size()) && A[negativeIdx] < 0)
         negativeIdx += 2;
      
      while ((positiveIdx < A.size()) && A[positiveIdx] > 0 )
         positiveIdx += 2;
      
      if (negativeIdx < A.size() && positiveIdx < A.size())
      {
         int tmp = A[negativeIdx];
         A[negativeIdx] = A[positiveIdx];
         A[positiveIdx] = tmp;
      }
   }
}
};