H-Index II -- leetcode
2015-10-05 15:08
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Follow up for H-Index:
What if the
is sorted in ascending order? Could you optimize your algorithm?
基本思路:折半查找
1. 选取中间点,判断该处是否满足hIndex定义
2. 如果满足定义,则在左区间继续搜索,以找到更大的hIndex
3. 如果不满足定义,在则在右区间搜索,以找到满足定义的点。
4. 重复步骤1.2.3,直到区间为空。
关于返回值设为 M - stop -1的正确性:
在缩小区间过程中,总是维持一不变式: stop后面的元素总是满足hIndex定义。
包括其初始值。即citations末尾之后的元素,也满足hIndex定义,其值为0。
class Solution {
public:
int hIndex(vector<int>& citations) {
const int M = citations.size();
int start = 0, stop = M - 1;
int hIndex = 0;
while (start <= stop) {
const int mid = start + (stop - start >> 1);
if (citations[mid] >= M - mid)
stop = mid - 1;
else
start = mid + 1;
}
return M - stop - 1;
}
};
What if the
citationsarray
is sorted in ascending order? Could you optimize your algorithm?
基本思路:折半查找
1. 选取中间点,判断该处是否满足hIndex定义
2. 如果满足定义,则在左区间继续搜索,以找到更大的hIndex
3. 如果不满足定义,在则在右区间搜索,以找到满足定义的点。
4. 重复步骤1.2.3,直到区间为空。
关于返回值设为 M - stop -1的正确性:
在缩小区间过程中,总是维持一不变式: stop后面的元素总是满足hIndex定义。
包括其初始值。即citations末尾之后的元素,也满足hIndex定义,其值为0。
class Solution {
public:
int hIndex(vector<int>& citations) {
const int M = citations.size();
int start = 0, stop = M - 1;
int hIndex = 0;
while (start <= stop) {
const int mid = start + (stop - start >> 1);
if (citations[mid] >= M - mid)
stop = mid - 1;
else
start = mid + 1;
}
return M - stop - 1;
}
};
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