LeetCode（35） Search Insert Position

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

分析

此题目与上一个题目类似都是源于二分查找的变形。
若序列中存在目标元素值，则直接返回其下标，若不存在则返回第一个大于它的元素的下标，只需要在二分搜索算法中稍微修改即可。

AC代码

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

    int BinarySearch(vector<int> & nums, int target)
    {
        int left = 0, right = nums.size() - 1;

        while (left <= right)
        {
            int mid = (left + right) / 2;
            if (nums[mid] == target)
                return mid;
            else if (nums[mid] < target)
            {
                if (mid == right || nums[mid + 1] > target)
                    return mid + 1;
                else
                    left = mid + 1;
            }
            else{
                if (mid == left || nums[mid - 1] < target)
                    return mid;
                else
                    right = mid - 1;
            }
        }//while

        return -1;
    }
};

GitHub测试程序源码