Difference between revisions of "TADM2E 4.34"
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return 1; | return 1; | ||
// Trivial case 2: elements in the array are continuous from 1 to n, the smallest missing integer must be n + 1 | // Trivial case 2: elements in the array are continuous from 1 to n, the smallest missing integer must be n + 1 | ||
− | else if(A[n - 1 | + | else if(A[n - 1] == n) |
{ | { | ||
return n + 1; | return n + 1; |
Revision as of 00:27, 23 September 2016
Assume array indices are zero based, use modified binary search to find the smallest integer missing:
FindMinIntMissing(A): // Trivial case 1 if(A[0] > 1) return 1; // Trivial case 2: elements in the array are continuous from 1 to n, the smallest missing integer must be n + 1 else if(A[n - 1] == n) { return n + 1; } // there is a gap in A, find the gap that is smallest else { return findGap(A, 0, n - 1); } FindGap(A, first, last): // only two elements left, the smallest gap must be the integer right after A[first] if(last == first + 1) return A[first] + 1; // recursively looking for the partition containing the smallest gap else mid = (first + last) / 2; // first half doesn't have gap, looking for gap in the second half if(A[mid] - A[first] == mid - first) return findGap(A, mid, last); // looking for the gap in the first half else return findGap(A, first, mid);