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
// Need to pay attention to the third example. nums.size() is included.
#include <vector>
#include <iostream>
using namespace std;
int searchIndex(vector<int>& nums, int target) {
if(nums.size() == 1) return 0;
int left = 0, right = nums.size(); // pay attention here!
while(left < right) {
int mid = (left + right) / 2;
if(nums[mid] >= target) right = mid;
else left = mid + 1;
}
return left;
}
int main(void) {
vector<int> nums{1, 3, 5, 6};
int pos = searchIndex(nums, 7);
cout << pos << endl;
}

本文介绍了一种在已排序数组中查找指定目标值的方法。若找到目标值,则返回其索引;若未找到,则返回目标值按序插入的位置。文章通过示例说明了算法的应用,并提供了完整的C++代码实现。
458

被折叠的 条评论
为什么被折叠?



