剑指Offer : 数组中的逆序对

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题目描述

在数组中的两个数字，如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组,求出这个数组中的逆序对的总数P。并将P对1000000007取模的结果输出。即输出P%1000000007

输入描述:

题目保证输入的数组中没有的相同的数字

数据范围：
对于%50的数据,size<=10^4
对于%75的数据,size<=10^5
对于%100的数据,size<=2*10^5

输入例子:
1,2,3,4,5,6,7,0

输出例子:

算法基于分治，归并排序的思想

代码如下：

typedef long long int64;class Solution{public:<span style="white-space:pre"></span>int InversePairs(vector<int> data)<span style="white-space:pre"></span>{<span style="white-space:pre"></span>int size = data.size();<span style="white-space:pre"></span>vector<int> copy(size);<span style="white-space:pre"></span>int64 res = merge(data, copy, 0, data.size() - 1);<span style="white-space:pre"></span>return res % 1000000007;<span style="white-space:pre"></span>}private:<span style="white-space:pre"></span>int64 merge(vector<int> & sourceArr, vector<int> & tempArr, int start, int end)<span style="white-space:pre"></span>{<span style="white-space:pre"></span>int64 res = 0;<span style="white-space:pre"></span>if(start < end){<span style="white-space:pre"></span>int middle = (start + end) >> 1;<span style="white-space:pre"></span>res += merge(sourceArr, tempArr, start, middle);<span style="white-space:pre"></span>res += merge(sourceArr, tempArr, middle + 1, end);<span style="white-space:pre"></span>res += merge_helper(sourceArr, tempArr, start, middle, end);<span style="white-space:pre"></span>}<span style="white-space:pre"></span>return res;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>int64 merge_helper(vector<int> & sourceArr, vector<int> & tempArr, int start, int middle, int end)<span style="white-space:pre"></span>{<span style="white-space:pre"></span>int64 count = 0;<span style="white-space:pre"></span>int i = middle, j = end, k = end;<span style="white-space:pre"></span>// <span style="white-space:pre"></span> i              j<span style="white-space:pre"></span>// start ...... middle ....... end<span style="white-space:pre"></span>while(i >= start && j >= middle + 1)<span style="white-space:pre"></span>{<span style="white-space:pre"></span>if(sourceArr[i] > sourceArr[j]){<span style="white-space:pre"></span>count += j - middle;<span style="white-space:pre"></span>tempArr[k--] = sourceArr[i--];<span style="white-space:pre"></span>}else{<span style="white-space:pre"></span>tempArr[k--] = sourceArr[j--];<span style="white-space:pre"></span>}<span style="white-space:pre"></span>}<span style="white-space:pre"></span>while(i >= start) <span style="white-space:pre"></span>tempArr[k--] = sourceArr[i--];<span style="white-space:pre"></span>while(j >= middle + 1) <span style="white-space:pre"></span>tempArr[k--] = sourceArr[j--];<span style="white-space:pre"></span>for(i = k + 1; i <= end; ++i){<span style="white-space:pre"></span>sourceArr[i] = tempArr[i];<span style="white-space:pre"></span>}<span style="white-space:pre"></span>return count;<span style="white-space:pre"></span>}};

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