629. K Inverse Pairs Array
2017-07-09 21:13
169 查看
题目:
Given two integers
find how many different arrays consist of numbers from
that there are exactly
We define an inverse pair as following: For
in the array, if
it's an inverse pair; Otherwise, it's not.
Since the answer may be very large, the answer should be modulo 109 + 7.
大概是求从1到n的各种组合中逆序数为k的个数
大神的做法:
dp
[k+1] = dp
[k]+dp[n-1][k+1]-dp[n-1][k+1-n]
在LeetCode上这道题已经有解析了,以下说一下自己的理解(大神请略过),自己在思考时,只考虑dp
[k]=d[n-1][k]+d[n-1][k-1]+....+dp[n-1][k-n+1]。然后,按照递归的for循环做,结果提示超时。按大神的思路,出现这种多个多项式相加的情况,我们可以再写另一个dp[n2][k2]使他们相减,然后消去相同的大量多项式。由于大量多项式是在向前n-1个数插入n,凑齐k时出现的,所以可以另n2=n,k2=k+1;然后相减,得到上述的等式。(在利用上述等式,利用嵌套加数组的形式解决时,也提示超时);
解读:
因为dp[i][j]-=dp[i-1][j-i],可能为负数,因为他们都是%mod的(如果未模除,前者一定比后者大),所以前者的大小未必比后者大,当前者较小时,我们可以加上mod,再进行模除。
Given two integers
nand
k,
find how many different arrays consist of numbers from
1to
nsuch
that there are exactly
kinverse pairs.
We define an inverse pair as following: For
ithand
jthelement
in the array, if
i<
jand
a[i]>
a[j]then
it's an inverse pair; Otherwise, it's not.
Since the answer may be very large, the answer should be modulo 109 + 7.
大概是求从1到n的各种组合中逆序数为k的个数
大神的做法:
dp
[k+1] = dp
[k]+dp[n-1][k+1]-dp[n-1][k+1-n]
public static int kInversePairs(int n, int k) { int mod = 1000000007; if (k > n*(n-1)/2 || k < 0) return 0; if (k == 0 || k == n*(n-1)/2) return 1; long[][] dp = new long[n+1][k+1]; dp[2][0] = 1; dp[2][1] = 1; for (int i = 3; i <= n; i++) { dp[i][0] = 1; for (int j = 1; j <= Math.min(k, i*(i-1)/2); j++) { dp[i][j] = dp[i][j-1] + dp[i-1][j]; if (j >= i) dp[i][j] -= dp[i-1][j-i]; dp[i][j] = (dp[i][j]+mod) % mod; } } return (int) dp [k]; }
在LeetCode上这道题已经有解析了,以下说一下自己的理解(大神请略过),自己在思考时,只考虑dp
[k]=d[n-1][k]+d[n-1][k-1]+....+dp[n-1][k-n+1]。然后,按照递归的for循环做,结果提示超时。按大神的思路,出现这种多个多项式相加的情况,我们可以再写另一个dp[n2][k2]使他们相减,然后消去相同的大量多项式。由于大量多项式是在向前n-1个数插入n,凑齐k时出现的,所以可以另n2=n,k2=k+1;然后相减,得到上述的等式。(在利用上述等式,利用嵌套加数组的形式解决时,也提示超时);
解读:
dp[i][j] = (dp[i][j]+mod) % mod;
因为dp[i][j]-=dp[i-1][j-i],可能为负数,因为他们都是%mod的(如果未模除,前者一定比后者大),所以前者的大小未必比后者大,当前者较小时,我们可以加上mod,再进行模除。
相关文章推荐
- 629. K Inverse Pairs Array
- K Inverse Pairs Array - LeetCode
- leetcode 629. K Inverse Pairs Array
- leetcode:K inverse pairs array
- 629. K Inverse Pairs Array 自制答案
- 629. K Inverse Pairs Array
- K Inverse Pairs Array (leetcode)
- [LeetCode] K Inverse Pairs Array K个翻转对数组
- 629. K Inverse Pairs Array
- [Leetcode] 629. K Inverse Pairs Array 解题报告
- Leetcode | K Inverse Pairs Array
- [leetcode]629. K Inverse Pairs Array
- K Inverse Pairs Array
- Leetcode之K-diff Pairs in an Array 问题
- [LeetCode] 532. K-diff Pairs in an Array
- K-diff Pairs In An Array
- Leetcode 532. K-diff Pairs in an Array
- K-diff Pairs in an Array
- 2017.11.22 LeetCode - 532. K-diff Pairs in an Array【简单二分】
- [LeetCode] K-diff Pairs in an Array 数组中差为K的数对