【LeetCode从零单刷】Search Insert Position
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题目:
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
解答:
题目不难,但是要注意边界条件:
- target value 小于或者等于 A[0] ;
- A 是个空数组;
- target value 大于 A 中所有元素;
- 除此之外,target value 是等于 A[i],还是等于A[i+1],还是(A[i], A[i+1])?
class Solution {public: int searchInsert(int A[], int n, int target) { if(n == 0 || A[0] >= target) return 0; if(A[n-1] < target) return n; int ans; for(int i = 0; i< n - 1; i++) { if(A[i] == target) { ans = i; } if(A[i+1] == target) { ans = (i+1); } if(A[i] < target && A[i+1] > target) { ans = (i+1); } } return ans; }};
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