35. Search Insert Position

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

题目含义：给定排序的数组和目标值，如果找到目标，返回索引。如果没有，就返回查找失败的当前索引

思想：

由于是已经排好序的，首先我想到的是for循环遍历，判断若目标值小于等于当前索引所在的值就返回索引值。时间是o(n)

还有一个方法，二分法，时间o(logn) 看了别人的代码才想到

C++ AC代码： 时间 o(n)

class Solution {public:    int searchInsert(vector<int>& nums, int target) {        int len = nums.size();        int index = 0;        while(index<len){            if(nums[index]>=target){                break;            }            index++;        }        return index;    }};

别人代码: 时间 o(logn)

class Solution {public:    int searchInsert(vector<int>& nums, int target) {        int l = 0;        int r = nums.size() - 1;        while (l <= r) {            int m = l + ((r - l) >> 1);             if (nums[m] == target) {                return m;            } else if (nums[m] < target) {                l = m + 1;            } else {                r = m - 1;            }        }        return l;    }};