本文地址: http://blog.csdn.net/caroline_wendy
排序的数组有可能包含正负, 可以使用折半查找确定中值位置, 然后再使用两个指针, 顺次排序两端.
解决思路是:
1.数组中的元素全为正,返回;
2.数组中的元素全为负,返回;
3.数组中有正数有负数,就用二分法查找,判断中间元素的符号
a)中间元素为正,继续判断中间元素前面一个元素的符号;
b)中间元素为负,判断中间元素后一个元素的符号;
c)中间元素为零,令其等于结果值返回;
时间复杂度O(nlogn)
代码:
/* * main.cpp * * Created on: 2014.9.12 * Author: Spike */ /*eclipse cdt, gcc 4.8.1*/ #include <stdio.h> #include <stdlib.h> using namespace std; void Reverse(int* begin, int* end) { while (begin < end) { int tmp = *begin; *begin++ = *end; *end-- = tmp; } } void AbsoluteSort(int res[], int data[], const int length) { if (data == NULL || length <= 0) { res = data; return; } if ((data[0]<0 && data[length-1]<0) || (data[0]>0 && data[length-1]>0)) { res = data; return; } int left = 0; int right = length-1; int mid; while (left < right) { mid = left + ((right-left)>>1); if (data[mid]>0 && data[mid-1]<=0) { right = mid; left = mid-1; break; } else if (data[mid]>0) { right = mid; } else if (data[mid]<0 && data[mid+1]>=0) { left = mid; right = mid+1; break; } else if (data[mid]<0) { left = mid; } else { break; } } int num=0; while (right < length && left >=0) { if (abs(data[right]) <= abs(data[left])) { res[num++] = data[right++]; } else { res[num++] = data[left--]; } } while (right < length) res[num++] = data[right++]; while (left >=0 ) { res[num++] = data[left--]; } } int main (void) { int data[] = {-5, -4, -2, -1, 0, 1, 2, 6}; const int length = sizeof(data)/sizeof(data[0]); int res[length]; AbsoluteSort(res, data, length); for (int i=0; i<length; ++i) { printf("%d ", res[i]); } printf("\n"); }
输出:
0 1 -1 2 -2 -4 -5 6
原文地址:http://blog.csdn.net/caroline_wendy/article/details/39273129