1. 题目
稀疏数组搜索。有个排好序的字符串数组,其中散布着一些空字符串,编写一种方法,找出给定字符串的位置。
示例1:
输入: words = ["at", "", "", "", "ball", "", "", "car", "", "","dad", "", ""], s = "ta"
输出:-1
说明: 不存在返回-1。
示例2:
输入:words = ["at", "", "", "", "ball", "", "", "car", "", "","dad", "", ""], s = "ball"
输出:4
提示:
words的长度在[1, 1000000]之间
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/sparse-array-search-lcci
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
2. 解题
- 将非空字符串和其序号,存起来
- 二分查找字符串,返回其序号
class Solution {
public:
int findString(vector<string>& words, string s) {
vector<pair<string,int>> w_id;
for(int i = 0; i < words.size(); ++i)
if(words[i] != "")
w_id.push_back({words[i],i});
int l = 0, r = w_id.size()-1, mid;
while(l <= r)
{
mid = l+((r-l)>>1);
if(s < w_id[mid].first)
r = mid-1;
else if(s > w_id[mid].first)
l = mid+1;
else
return w_id[mid].second;
}
return -1;
}
};
- 上面画蛇添足了,直接遍历过一遍了,还二分查找。。
- 直接二分查找如下:需要对空字符串进行处理跳过,mid 也是挪至非空处
class Solution {
public:
int findString(vector<string>& words, string s) {
int l = 0, r = words.size()-1, mid;
while(l <= r)
{
while(l <= r && words[l]=="")
l++;
while(l <= r && words[r]=="")
r--;
mid = l+((r-l)>>1);
while(l <= mid && words[mid]=="")
mid--;
// while(r >= mid && words[mid]=="") //这样写也行
// mid++;
if(s < words[mid])
r = mid-1;
else if(s > words[mid])
l = mid+1;
else
return mid;
}
return -1;
}
};