LeetCode 435. Non-overlapping Intervals 无重叠区间(贪心问题，两种解法)
LeetCode 435. Non-overlapping Intervals 无重叠区间
- 435. Non-overlapping Intervals 无重叠区间
- 题目描述
- 示例:
- 解答1
- 代码1
- 解答1
- 代码1
435. Non-overlapping Intervals 无重叠区间
题目描述
Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Note:
You may assume the interval’s end point is always bigger than its start point.
Intervals like [1,2] and [2,3] have borders “touching” but they don’t overlap each other.
示例:
Example 1:
Input: [ [1,2], [2,3], [3,4], [1,3] ]
Output: 1
Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Example 2:
Input: [ [1,2], [1,2], [1,2] ]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Example 3:
Input: [ [1,2], [2,3] ]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
解答1
这道题和用最少数量的箭引爆气球类似，都是贪心算法。
第一种思路，我们可以先按照interval的右坐标进行从小到大排序。然后从第二个interval开始，每当第二个interval的左坐标小于第一个interval的右坐标，我们令count++。
代码1
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
if (intervals.empty()) return 0;
sort(intervals.begin(), intervals.end(),[&](const vector<int >&a, const vector<int> &b){return a[1] < b[1];});
//本来count应该初始化为0，但是因为底下的for循环，又把intervals[0][1]考虑了一遍
//所以这里count 初始化为0 - 1
auto count = -1;
auto lastPoint = intervals[0][1];
for (auto & i : intervals) {
if (i[0] < lastPoint) {
count++;
if (i[1] < lastPoint) lastPoint = i[1];
}
else
lastPoint = i[1];
}
return count;
}
};
解答1
第二种思路，找到需要去掉的最少interval的数量，就是总数量减去不重叠interval的数量。
代码如下。
代码1
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
if (intervals.empty()) return 0;
sort(intervals.begin(), intervals.end(),[&](const vector<int >&a, const vector<int> &b){return a[1] < b[1];});
auto count = 1;
auto lastPoint = intervals[0][1];
for (auto & i : intervals) {
if (i[0] >= lastPoint) {
count++;
lastPoint = i[1];
}
}
return intervals.size() - count;
}
};