算法（十七）贪心问题

题目描述

An integer interval [a, b] (for integers a < b) is a set of all consecutive integers from a to b, including a and b.

Find the minimum size of a set S such that for every integer interval A in intervals, the intersection of S with A has size at least 2.

Example 1:
Input: intervals = [[1, 3], [1, 4], [2, 5], [3, 5]]
Output: 3
Explanation:
Consider the set S = {2, 3, 4}.  For each interval, there are at least 2 elements from S in the interval.
Also, there isn't a smaller size set that fulfills the above condition.
Thus, we output the size of this set, which is 3.

Example 2:
Input: intervals = [[1, 2], [2, 3], [2, 4], [4, 5]]
Output: 5
Explanation:
An example of a minimum sized set is {1, 2, 3, 4, 5}.

解题思路

将每个[a,b]值按b的值从大到小排列，然后循环每个[a,b]值，这时候有三种情况，下一个对的最小值小于上一个对的b值，所以不需要计算。下一个对的最小值大于上一个对的b值，总数加2，如果等于的话总数加1。

代码解析：

class Solution {
public:
int intersectionSizeTwo(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), [](vector<int>& a, vector<int>& b) {
return a[1] < b[1] || (a[1] == b[1] && a[0] > b[0]);
});
int n = intervals.size(), ans = 0, p1 = -1, p2 = -1;
for (int i = 0; i < n; i++) {
// current p1, p2 works for intervals[i]
if (intervals[i][0] <= p1) continue;
// Neither of p1, p2 works for intervals[i]
// replace p1, p2 by ending numbers
if (intervals[i][0] > p2) {
ans += 2;
p2 = intervals[i][1];
p1 = p2-1;
}
// only p2 works;
else {
ans++;
p1 = p2;
p2 = intervals[i][1];
}
}
return ans;
}
};