Magic index

find the index with value equal to index. Time Complexity: O(logn), Space Complexity: O(1).

//numbers are distinctpublic int getMagic(int[] a, int start, int end) {if(start > end) return -1;int middle = start + (end - start) / 2;if(a[middle] == middle) return middle;else if(a[middle] > middle) return getMagic(a, start, middle - 1);else return getMagic(a, middle + 1, end);}//if numbers are not distinct, means have repeated numberspublic int getMagic2(int[] a, int start, int end) {if(start > end) return -1;int middle = start + (end - start) / 2;if(a[middle] == middle) return middle;//search left//the min leftIndex should be compare with the value int leftIndex = Math.min(middle - 1, a[middle]);int left = getMagic2(a, start, leftIndex);if(left >= 0) return left;//search right int rightIndex = Math.max(middle + 1, a[middle]);int right = getMagic2(a, rightIndex, end);return right;}