leetcode35

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

解法一：
遍历一遍原数组，若当前数字大于或等于目标值，则返回当前坐标，如果遍历结束了，说明目标值比数组中任何一个数都要大，则返回数组长度n即可。

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

解法二：
用二分搜索法来优化我们的时间复杂度。

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