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Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume NO duplicates in the array.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0
O(log(n)) time
Solution 1
Naively, we can just iterate the array and compare target with ith and (i+1)th element. Time complexity is O(n)
1 public int searchInsert(int[] A, int target) 2 { 3 // write your code here 4 if(A.length==0) return 0; 5 6 if(target <= A[0]) return 0; 7 8 for(int i=0; i<A.length-1; i++){ 9 if(target > A[i] && target <= A[i+1]){ 10 return i+1; 11 } 12 } 13 14 return A.length; 15 16 }
Solution 2
This also looks like a binary search problem. We should try to make the complexity to be O(log(n)).
1 public int searchInsert(int[] A, int target) 2 { 3 // write your code here 4 if(A.length==0) return 0; 5 return helper(A, target, 0, A.length-1); 6 7 } 8 9 10 public int helper(int[] A, int target, int start, int end) 11 { 12 int mid = (start+end)/2; 13 14 if (target == A[mid]) return mid; 15 16 if(target<A[mid]) 17 { 18 if (mid <= start) return start; 19 else 20 { 21 return helper(A,target,start,mid-1); 22 } 23 } 24 25 else //target > A[mid] 26 { 27 if(mid >= end) return end+1; 28 else 29 { 30 return helper(A,target,mid+1,end); 31 } 32 } 33 34 35 }
reference: http://www.programcreek.com/2013/01/leetcode-search-insert-position/
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原文地址：http://www.cnblogs.com/hygeia/p/4779852.html
