[LintCode] Reverse Pairs

tigerZH 發(fā)布于2019-08-14 15:33 / 877人閱讀

摘要：暴力解法就是時(shí)靈時(shí)不靈，兩次一次。希望看到的大神能夠分享優(yōu)質(zhì)的解法謝謝大家

Problem

For an array A, if i < j, and A[i] > A[j], called (A[i], A[j]) is a reverse pair.
return total of reverse pairs in A.

Example

Given A = [2, 4, 1, 3, 5] , (2, 1), (4, 1), (4, 3) are reverse pairs. return 3

Note

暴力解法就是時(shí)靈時(shí)不靈，submit兩次ac一次。
希望看到的大神能夠分享優(yōu)質(zhì)的解法^_^ linspiration1991@gmail.com

Solution

public class Solution {
    public long reversePairs(int[] A) {
        long res = 0;
        int n = A.length;
        for (int i = 0; i < n-1; i++) {
            for (int j = i; j < n; j++) {
                if (A[i] > A[j]) res += 1;
            }
        }
        return res;
    }
}
//謝謝大家！

GPU云服務(wù)器云服務(wù)器 pairs LintCode Reverse LintCode134_LRU

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