标签：pairs seq wro 存储 images lead tps code 存储方式
my solution:
class Solution { public: int reversePairs(vector<int>& nums) { int length=nums.size(); int count=0; for (int i=0;i<length;i++) { for(int j=i+1;j<length;j++) { if(nums[i]>2*nums[j]) count++; } } return count; } };
wrong answer ：
because 2147483647*2 is -2 (int) , only 33bit can represent
| int | 4 byte | 32 bit |
2147483647--2147483647 signed 0-4294967295 unsigned |
| long | 4 byte | 32 bit | （在 32 位机器上与int 相同） |
| double | 8 byte | 64 bit | 1.79769e+308 ~ 2.22507e-308 |
| float | 4 byte | 32 bit | 3.40282e+038 ~ 1.17549e-038 |
| long double | 12 byte | 96 bit | 1.18973e+4932 ~ 3.3621e-4932 |
then i change the type to double:
class Solution { public: int reversePairs(vector<int>& nums) { int length=nums.size(); int count=0; double a; for (int i=0;i<length;i++) { for(int j=i+1;j<length;j++) { a=(double)2*nums[j]; if(nums[i]>a) count++; } } return count; } };
wrong:
because the if the maxmise length is 50000, the process will lead to time exceed .
对于这类问题，一种很好的解法就是拆分数组来解决子问题，通过求解小问题来求解大问题。
有两类拆分方法：
class Solution { public: int reversePairs(vector<int>& nums) { return mergeSort(nums, 0, nums.size() - 1); } int mergeSort(vector<int>& nums, int left, int right) { if (left >= right) return 0; int mid = left + (right - left) / 2; int res = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right); for (int i = left, j = mid + 1; i <= mid; ++i) { while (j <= right && nums[i] / 2.0 > nums[j]) ++j; res += j - (mid + 1); } sort(nums.begin() + left, nums.begin() + right + 1); return res; } };
归并排序 MergeSort 和BIT可以解决，BST和 binary search不行
https://discuss.leetcode.com/topic/79227/general-principles-behind-problems-similar-to-reverse-pairsBST （binary search tree）
BIT （binary indexed tree）
标签：pairs seq wro 存储 images lead tps code 存储方式
原文地址：http://www.cnblogs.com/fanhaha/p/7142651.html
