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求解最長遞增子序列長度|動態規劃+二分查找:C\C++實現
01
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/*
求解最長遞增子序列長度 */
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02
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#include
<iostream>
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03
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#include
<limits>
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04
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05
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using namespace std;
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06
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07
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//int
x[] = {1,-1,2,-3,4,-5,6,-7};
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08
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int x[]
= {2, 1, 5, 3, 6, 4, 8, 9, 7};
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09
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int disp_array(int a[], int len);
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10
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int find_first_bigger(int a[], int last, int key);
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12
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int main()
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13
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{
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int len
= sizeof(x)/sizeof(x[0]);
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15
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int *z
= new int(len
+ 1);
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/*
z[i] = min(t|t:長度為i的子序列的最後一個元素) */
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z[0]
= std::numeric_limits<int>::min();
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for (int i
= 1; i <= len; ++i)
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{
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z[i]
= 0;
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}
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23
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disp_array(x,
len);
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disp_array(z,
len + 1);
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int t,
lislen = 0;
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for (int i
= 0; i < len; ++i)
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{
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t
= find_first_bigger(z, lislen, x[i]);
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if (t
> lislen)
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z[++lislen]
= x[i]; /*
z[]中沒有比x[i]大的*/
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else
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z[t]
= x[i]; /*
z[t]第一個比x[i] */
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}
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disp_array(z,
len + 1);
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cout
<< "LIS
= " <<
lislen << endl;
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return 0;
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}
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/**
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*
@brief 在數組a[0..last]中，自左向又查找第一個比key大的數的下標
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*
時間複雜度O(lg(n))
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*
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*
@param a[]
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*
@param last
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*
@param key
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*
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*
@return 返回下標，如果沒有比key大的則返回last+1
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*/
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int find_first_bigger(int a[], int last, int key)
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54
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{
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int m,
p = -1, q = last + 1;
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/*
assume: a[-1] < key <= a[last + 1] */
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while (p
+ 1 != q)
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{
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/*
invartant: 0 <= p + 1 < q <= last + 1 && a[p] < key <= a[q]*/
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m
= p + ((q - p)>>1);
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if (a[m]
< key)
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p
= m;
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else
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q
= m;
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}
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return q;
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}
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int disp_array(int a[], int len)
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71
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{
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for (int i
= 0; i < len; ++i)
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{
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cout
<< a[i];
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if (i
== len -1)
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cout
<< endl;
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else
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cout
<< "
";
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}
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81
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}
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最後更新：2017-04-03 22:15:45