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
本题利用二分查找的想法，寻找target的合适位置。时间：17ms。代码如下：
class Solution { public: int searchInsert(vector<int>& nums, int f, int l, int target) { if (nums[f] >= target) return f; int mid = f + (l - f) / 2; if (target <= nums[mid]) return searchInsert(nums, f, mid - 1, target); else return searchInsert(nums, mid + 1, l, target); } int searchInsert(vector<int>& nums, int target) { if ((nums.size() == 0) || (target <= nums.front())) return 0; if (target>nums.back()) return nums.size(); return searchInsert(nums, 0, nums.size() - 1, target); } };
[LeetCode] #35 Search Insert Position
原文：http://www.cnblogs.com/Scorpio989/p/4579287.html