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
思路:
这其实就是STL中lower_bound的实现:找到数组中第一个大于等于target的数字所对应的下标。完全可以参考Leetcode 34中相对应的解法。
代码:
class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
if(nums.size() == 0)
return 0;
int left = 0, right = nums.size() - 1;
while(left <= right)
{
int mid = left + (right - left) / 2;
if(nums[mid] >= target)
right = mid - 1;
else
left = mid + 1;
}
return left;
}
};
本文介绍了一种查找算法,用于确定目标值在已排序数组中的插入位置,确保数组仍保持有序状态。该算法类似于STL中的lower_bound,通过二分搜索找到首个大于等于目标值的元素下标。
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