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Copy path0035_search_insert_position.cpp
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81 lines (76 loc) · 2.43 KB
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/*
* Copyright(c) 2019 Jiau Zhang
* For more information see <https://github.com/JiauZhang/algorithms>
*
* This repo is free software: you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation
*
* It is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with THIS repo. If not, see <http://www.gnu.org/licenses/>.
*/
/*
* https://leetcode-cn.com/problems/search-insert-position
* 题目描述：
* 给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。
* 如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
* 你可以假设数组中无重复元素。
*
* 示例 1:
* 输入: [1,3,5,6], 5
* 输出: 2
*
* 示例 2:
* 输入: [1,3,5,6], 2
* 输出: 1
*
* 示例 3:
* 输入: [1,3,5,6], 7
* 输出: 4
*
* 示例 4:
* 输入: [1,3,5,6], 0
* 输出: 0
*
* 解题思路：
* 1. 最简单粗暴的方法就是逐个遍历，直到找到目标数字或者第一个
* 大于目标数的值，不过时间复杂度为 O(N)
* 2. 由于给定的数组是有序数组，所以可以使用二分法进行查找
* 这样时间复杂度就可以变成 O(logN)
*/
class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
if (nums.size() == 0)
return 0;
int start = 0, end = nums.size() - 1;
if (nums[start] > target)
return 0;
if (nums[end] < target)
return end + 1;
int mid;
while (start < end) {
/*
本来想优化一下，结果这句造成了死循环
按理说有符号的正数不应该出现这种情况的呀！
mid = start + (end - start) >> 1;
*/
mid = start + (end - start) / 2;
if (nums[mid] > target) {
end = mid - 1;
} else if (nums[mid] < target) {
start = mid + 1;
} else {
return mid;
}
}
if (nums[start] >= target)
return start;
return start + 1;
}
};