[Leetcode] 35. Search Insert Position 解题报告
2016-12-29 06:42
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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.
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:
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; } };
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