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.
Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0
普通方法，时间复杂度为O(n)
1 class Solution { 2 public: 3 int searchInsert(int A[], int n, int target) { 4 if(n==0) return NULL; 5 int i=0; 6 for(i=0;i<n;i++) 7 { 8 if(target<=A[i]) 9 return i; 10 } 11 return i; 12 } 13 };
二分法，时间复杂度为O(logn)
1 class Solution { 2 public: 3 int searchInsert(int A[], int n, int target) { 4 int start=0; 5 int end=n-1; 6 int index=(start+end)/2; 7 while(start<end) 8 { 9 if(A[index]>target) 10 { 11 end=index-1; 12 index=(start+end)/2; 13 }else if(A[index]<target){ 14 start=index+1; 15 index=(start+end)/2; 16 }else{ 17 return index; 18 } 19 } 20 if(A[start]<target){ 21 return start+1; 22 }else{ 23 return start; 24 } 25 } 26 };
