35. Search Insert Position

这两天需要先去读一些RFC啥的，没空做了。

Title

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

二分查找，然后插入，时间lg(n)。

Solution

one

class Solution {public:    int searchInsert(vector<int>& nums, int target) {        int n = nums.size();        int low = 0, high = n-1, mid=0;        while (low <= high) {            mid = low + (high-low)/2;            if (nums[mid] == target) {                return mid;            }            else if (nums[mid] < target) {                low = mid+1;            }            else {                high = mid-1;            }        }        if (nums[mid] > target) {            return mid;        }        return mid+1;    }};

runtime: 8ms

two

date: 2016.9.16

二分查找，直接找到lower_bound解决此问题，关于二分查找的总结，请看这里：二分查找

class Solution {public:    int searchInsert(vector<int>& nums, int target) {        int n = (int)nums.size();        int low = 0, high = n, mid;        while (low < high) { //输入[low, high); high == nums.size()            mid = low + (high-low)/2;            if (nums[mid] >= target) { //继续查找[low, mid]                high = mid;            }            else { //[mid+1, high]                low = mid+1;            }        }        return low; //返回[low, high]    }};

runtime: 6ms