Created
May 9, 2013 11:33
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Dynamic Programing : Longest Increasing Sequence (LIS)
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// DP version, Time Complexity : O(n^2) | |
public int lis(int[] A){ | |
if (A == null || A.length == 0) return 0; | |
int n = A.length; | |
int[] f = new int[n]; | |
for (int i=0; i<n; i++){ | |
f[i] = 1; | |
for (int j=0; j<i; j++){ | |
if (A[j] <= A[i] && f[j] + 1 > f[i]) | |
f[i] = f[j] + 1; | |
} | |
} | |
return f[n-1]; | |
} | |
// O(nlgn) version by using binary search | |
public int lis(int[] A){ | |
if (A == null || A.length == 0) return 0; | |
int[] d = new int[A.length]; | |
d[0] = A[0]; | |
k = 0; | |
for (int i=1; i<A.length; i++){ | |
int j = findFirstGreaterThanElement(d, 0, k, A[i]); | |
d[j] = A[i]; | |
if (d > k) | |
k++; | |
} | |
return k+1; | |
} | |
int findFirstGreaterThanElement(int[] A, int low, int high, int key){ | |
while (low <= high){ | |
int mid = low + (high - low)/2; | |
if (A[mid] <= key) | |
low = mid+1; | |
else | |
high = mid-1; | |
} | |
return low; | |
} |
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