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
解法1：顺序查找，时间复杂度O(n)。
class Solution { public: int searchInsert(vector<int>& nums, int target) { int n = nums.size(), i = 0; while(i < n && target > nums[i]) i++; return i; } };
解法2：因为是排序数组，必然想到二分查找，时间复杂度O(logn)。
class Solution { public: int searchInsert(vector<int>& nums, int target) { int n = nums.size(); if(n < 1) return 0; int index = n; int left = 0, right = n - 1; while(left < right) { int mid = (left + right) >> 1; if(target == nums[mid]) { index = mid; return index; } else if(target < nums[mid]) right = mid - 1; else left = mid + 1; } index = target <= nums[left] ? left : ++left; return index; } };
[LeetCode]24. Search Insert Position插入位置
原文：http://www.cnblogs.com/aprilcheny/p/4875709.html