给定一个区间的集合 intervals
,其中 intervals[i] = [starti, endi]
。返回 需要移除区间的最小数量,使剩余区间互不重叠 。
示例 1:
输入: intervals = [[1,2],[2,3],[3,4],[1,3]] 输出: 1 解释: 移除 [1,3] 后,剩下的区间没有重叠。
示例 2:
输入: intervals = [ [1,2], [1,2], [1,2] ] 输出: 2 解释: 你需要移除两个 [1,2] 来使剩下的区间没有重叠。
示例 3:
输入: intervals = [ [1,2], [2,3] ] 输出: 0 解释: 你不需要移除任何区间,因为它们已经是无重叠的了。
提示:
1 <= intervals.length <= 105
intervals[i].length == 2
-5 * 104 <= starti < endi <= 5 * 104
最长上升子序列问题,动态规划的做法,时间复杂度是
class Solution:
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
intervals.sort(key=lambda x: x[1])
ans, t = 0, intervals[0][1]
for s, e in intervals[1:]:
if s >= t:
t = e
else:
ans += 1
return ans
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
Arrays.sort(intervals, Comparator.comparingInt(a -> a[1]));
int t = intervals[0][1], ans = 0;
for (int i = 1; i < intervals.length; ++i) {
if (intervals[i][0] >= t) {
t = intervals[i][1];
} else {
++ans;
}
}
return ans;
}
}
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), [](const auto& a, const auto& b) { return a[1] < b[1]; });
int ans = 0, t = intervals[0][1];
for (int i = 1; i < intervals.size(); ++i) {
if (t <= intervals[i][0])
t = intervals[i][1];
else
++ans;
}
return ans;
}
};
func eraseOverlapIntervals(intervals [][]int) int {
sort.Slice(intervals, func(i, j int) bool {
return intervals[i][1] < intervals[j][1]
})
t, ans := intervals[0][1], 0
for i := 1; i < len(intervals); i++ {
if intervals[i][0] >= t {
t = intervals[i][1]
} else {
ans++
}
}
return ans
}
function eraseOverlapIntervals(intervals: number[][]): number {
intervals.sort((a, b) => a[1] - b[1]);
let end = intervals[0][1],
ans = 0;
for (let i = 1; i < intervals.length; ++i) {
let cur = intervals[i];
if (end > cur[0]) {
ans++;
} else {
end = cur[1];
}
}
return ans;
}
先按照区间右边界排序。优先选择最小的区间的右边界作为起始边界。遍历区间:
- 若当前区间左边界大于等于起始右边界,说明该区间无需移除,直接更新起始右边界;
- 否则说明该区间需要移除,更新移除区间的数量 ans。
最后返回 ans 即可。
时间复杂度
class Solution:
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
intervals.sort()
d = [intervals[0][1]]
for s, e in intervals[1:]:
if s >= d[-1]:
d.append(e)
else:
idx = bisect_left(d, s)
d[idx] = min(d[idx], e)
return len(intervals) - len(d)
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> {
if (a[0] != b[0]) {
return a[0] - b[0];
}
return a[1] - b[1];
});
int n = intervals.length;
int[] d = new int[n + 1];
d[1] = intervals[0][1];
int size = 1;
for (int i = 1; i < n; ++i) {
int s = intervals[i][0], e = intervals[i][1];
if (s >= d[size]) {
d[++size] = e;
} else {
int left = 1, right = size;
while (left < right) {
int mid = (left + right) >> 1;
if (d[mid] >= s) {
right = mid;
} else {
left = mid + 1;
}
}
d[left] = Math.min(d[left], e);
}
}
return n - size;
}
}