• 動態規劃
  $dp[i]$表示以區間$i$結尾，可以共存最多的區間數量。
  $dp[i]=max(dp[i],dp[j]+1)$，區間$j$滿足右邊界$\le$區間$i$的左邊界。

class Solution { public:int eraseOverlapIntervals(vector<vector<int>>& intervals) {int n = intervals.size();if (n < 2) {return 0;}sort(intervals.begin(), intervals.end(), [](vector<int> &l, vector<int> &r) {return l[0] < r[0];});vector<int> dp(n, 1);for (int i = 1; i < n; ++i) {for (int j = 0; j < i; ++j) {if (intervals[j][1] <= intervals[i][0]) {dp[i] = max(dp[i], dp[j] + 1);}}}return n - *max_element(dp.begin(), dp.end());} }; class Solution:def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:if not intervals:return 0intervals.sort()n = len(intervals)f = [1]for i in range(1, n):# 列表推導優化常熟f.append(max((f[j] for j in range(i) if intervals[j][1] <= intervals[i][0]), default=0) + 1)return n - max(f)

• 貪心
  區間按照右端點排序，每次貪心的找最左邊的區間，因為這就是最優的。

class Solution { public:int eraseOverlapIntervals(vector<vector<int>>& intervals) {int n = intervals.size();if (n < 2) {return 0;}sort(intervals.begin(), intervals.end(), [](vector<int> &l, vector<int> &r) {return l[1] < r[1];});int ret = 0;int pre = intervals[0][1];for (int i = 1; i < n; ++i) {if (intervals[i][0] < pre) {ret++;}else {pre = intervals[i][1];}}return ret;} }; class Solution:def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:if not intervals:return 0intervals.sort(key=lambda x: x[1])n = len(intervals)ret = 0right = intervals[0][1]for i in range(1, n):if intervals[i][0] < right:ret += 1else:right = intervals[i][1]return ret

總結

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