程序员面试金典 - 面试题 10.05. 稀疏数组搜索（二分查找）

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;
    }
};