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相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。仅仅须要在二分搜索算法中略微改动就可以。
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;
}
};