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

思路1：直觀的想法，遍歷數(shù)組，一旦當前元素>=target，當前元素的位置即為插入位置，邊界問題處理一下即可。時間復雜度：O(n)

class Solution { public:int searchInsert(int A[], int n, int target) {if(target > A[n-1])return n;else { for(int i = 0; i < n; i++) {if(A[i] >= target)return i;}}} };
思路2：由于給定數(shù)組為已經(jīng)排好序的，可以使用二分查找，比較A[mid]與target的大小。時間復雜度：O(logn)

class Solution { public:int searchInsert(int A[], int n, int target) {int start = 0;int end = n - 1;int mid = 0;while(end >= start) {mid = (start + end) / 2;if(target == A[mid])return mid;else if(target > A[mid])start = mid + 1;elseend = mid - 1;}if(target > A[mid])return mid + 1;elsereturn mid;} };

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