class Solution {
public:
bool canAttendMeetings(vector<vector<int>>& intervals) {
}
};
252. 会议室
给定一个会议时间安排的数组 intervals ，每个会议时间都会包括开始和结束的时间 intervals[i] = [starti, endi] ，请你判断一个人是否能够参加这里面的全部会议。
示例 1：
输入：intervals = [[0,30],[5,10],[15,20]] 输出：false
示例 2：
输入：intervals = [[7,10],[2,4]] 输出：true
提示：
0 <= intervals.length <= 104intervals[i].length == 20 <= starti < endi <= 106原站题解
golang 解法, 执行用时: 8 ms, 内存消耗: 4.1 MB, 提交时间: 2023-10-15 19:11:56
//其实就是判断两个区间是否相交，如果相交直接返回true
func canAttendMeetings(intervals [][]int) bool {
//对intervals进行排序:按照起点进行升序排序
sort.Slice(intervals, func(i, j int) bool {
return intervals[i][0] < intervals[j][0]
})
//之后遍历我们的intervals
for i := 1; i < len(intervals); i++ {
//如果当前区间与上一区间起点相同，或者当前区间的起点小于上一区间的末尾，直接返回false
if intervals[i][0] == intervals[i-1][0] || intervals[i][0] < intervals[i-1][1]{
return false
}
}
return true
}
python3 解法, 执行用时: 48 ms, 内存消耗: 18.3 MB, 提交时间: 2023-10-15 19:11:14
class Solution:
def canAttendMeetings1(self, intervals: List[List[int]]) -> bool:
def overlap(interval1: List[int], interval2: List[int]) -> bool:
return (interval1[0] >= interval2[0] and interval1[0] < interval2[1]
or interval2[0] >= interval1[0] and interval2[0] < interval1[1])
for i in range(len(intervals)):
for j in range(i + 1, len(intervals)):
if overlap(intervals[i], intervals[j]):
return False
return True
# 排序
def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
intervals.sort()
for i in range(len(intervals) - 1):
if intervals[i][1] > intervals[i + 1][0]:
return False
return True
java 解法, 执行用时: 4 ms, 内存消耗: 43.4 MB, 提交时间: 2023-10-15 19:10:42
class Solution {
// 暴力
public boolean canAttendMeetings1(int[][] intervals) {
for (int i = 0; i < intervals.length; i++) {
for (int j = i + 1; j < intervals.length; j++) {
if (overlap(intervals[i], intervals[j])) {
return false;
}
}
}
return true;
}
private boolean overlap(int[] interval1, int[] interval2) {
return (interval1[0] >= interval2[0] && interval1[0] < interval2[1])
|| (interval2[0] >= interval1[0] && interval2[0] < interval1[1]);
}
// 排序
public boolean canAttendMeetings(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
for (int i = 0; i < intervals.length - 1; i++) {
if (intervals[i][1] > intervals[i + 1][0]) {
return false;
}
}
return true;
}
}
cpp 解法, 执行用时: 12 ms, 内存消耗: 11.7 MB, 提交时间: 2023-10-15 19:10:10
class Solution {
public:
bool overlap(vector<int>& interval1, vector<int>& interval2) {
return interval1[0] >= interval2[0] and interval1[0] < interval2[1]
or interval2[0] >= interval1[0] and interval2[0] < interval1[1];
}
// 暴力
bool canAttendMeetings1(vector<vector<int>>& intervals) {
for (size_t i = 0; i < intervals.size(); i++) {
for (size_t j = i + 1; j < intervals.size(); j++) {
if (overlap(intervals[i], intervals[j])) {
return false;
}
}
}
return true;
}
// 排序
bool canAttendMeetings(vector<vector<int>>& intervals) {
if (intervals.empty()) {
return true;
}
// 注意：C++ 排序函数会自动首先按第一个元素对向量排序，然后按第二个元素排序，依此类推。
sort(intervals.begin(), intervals.end());
for (size_t i = 0; i < intervals.size() - 1; i++) {
if (intervals[i][1] > intervals[i + 1][0]) {
return false;
}
}
return true;
}
};