Algorithm about Subarrays & substrings

The Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which have the largest sum.

e.g. [−2,1,−3,4,−1,2,1,−5,4] -> [4,−1,2,1] = 6 .

1  Public voidMaxsubsimple (int[] Array) {2     if(Array = =NULL|| Array.Length = = 0) {3System.out.println ("Empty Array");4     }5     intmax = Array[0], cursum = array[0];6      for(inti = 1; i < Array.Length; i++) {7Cursum = cursum > 0? Cursum +Array[i]: array[i];8max = max > cursum?max:cursum;9     }TenSystem.out.println ("Max Sub-array sum is:" +max); One } A  Public voidMaxsubfull (int[] Array) { -     if(Array = =NULL|| Array.Length = = 0) { -System.out.println ("Empty Array"); the     } -     intmax = Array[0], cursum = array[0]; -     intStart = 0, end = 0; -     intfinal_s = 0, final_t = 0; +      for(inti = 1; i < Array.Length; i++) { -         if(Cursum < 0) { +Cursum =Array[i]; AStart =i; atEnd =i; -}Else { -Cursum = Cursum +Array[i]; -End =i; -         } -         if(Max <cursum) { inMax =cursum; -final_s =start; tofinal_t =end; +         } -     } theSystem.out.println ("Sub Array: (" +final_s+ "," +final_t+ ")"); *System.out.println ("Max Sub-array sum is:" +max); $}

View Code

No.2 Maximum Product Subarray

Find the contiguous subarray within an array (containing at least one number) which have the largest product.

e.g. [2,3,-2,4] -> [2,3] = 6 .

1  Public intMaxproduct (int[] A) {2         intPremin = A[0], Premax = a[0], max = a[0];3          for(inti = 1; i < a.length; i++) {4             intTMP1 = Premin *A[i];5             intTMP2 = Premax *A[i];6Premin =math.min (A[i], math.min (TMP1, TMP2));7Premax =Math.max (A[i], Math.max (TMP1, TMP2));8Max =Math.max (Premax, max);9         }Ten         returnMax; One}

View Code

No.3 LIS (longest increasing subarray)

The increasing subarray does ' t have to be contiuous nor same diff

e.g. [1, 6, 8, 3, 7, 9], [1, 3, 7, 9] = 4

DP, O (n^2)

1  Public voidFindlis (int[] Array) {2     if(Array = =NULL|| Array.Length = = 0) {3System.out.println ("Empty array");4         return;5     }6     intLen =Array.Length;7     int[] DP =New int[Len];8     intLis = 1;9      for(inti = 0; i < Len; i++) {TenDp[i] = 1; One          for(intj = 0; J < I; J + +) { A             if(Array[j] < Array[i] & Dp[j] + 1 >Dp[i]) { -Dp[i] = Dp[j] + 1; -             } the         } -         if(Dp[i] >lis) { -Lis =Dp[i]; -         } +     } -System.out.println ("Length of longest increasing subarray:" +lis); +}

View Code

DP + Binary Search, O (NLOGN)

Maintain an array maxval[i], record the minimum value of the largest element in the increment subsequence of length I, and for each element in the array, examine the largest element of which subsequence it is, and binary update maxval array-the beauty of programming

1  Public voidBinsearch (int[] Maxval,intMaxLen,intx) {2     intleft = 0, right = maxlen-1;3      while(Left <=Right ) {4         intMid = left + (right-left)/2;5         if(Maxval[mid] <=x) {6left++;7}Else {8right--;9         }Ten     } OneMaxval[left] =x; A } -  Public voidLIS (int[] Array) { -     if(Array = =NULL|| Array.Length = = 0) { theSystem.out.println ("Empty array"); -         return; -     } -     intLen =Array.Length; +     int[] Maxval =New int[Len]; -Maxval[0] = array[0]; +     intMaxLen = 1; A      for(inti = 0; i < Len; i++) { at         if(Array[i] > maxval[maxlen-1]) { -maxval[maxlen++] =Array[i]; -}Else { - Binsearch (Maxval, MaxLen, Array[i]); -         } -     } inSystem.out.println ("Length of longest increasing subarray:" +maxlen); -}

View Code

No.4 LCS (Longest Common Subarray)

Algorithm about Subarrays & substrings