LeetCode刷題之貪心演算法—重疊區間
阿新 • • 發佈:2022-04-05
1.重疊區間:一組二維陣列,它的0列是開始,1列是結束。最少要刪除多少個子陣列,各子區間才不重疊
方法一:直接記錄交叉區間個數
static bool cmp(vector<int>& a, vector<int>& b)
{
return a[1]<b[1];//右列排序
}
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
if(intervals.size()==0) return 0;
sort(intervals.begin(),intervals.end(),cmp);
int res=0;//交叉數
for(int i=1; i<intervals.size();i++)
{
if(intervals[i][0]<intervals[i-1][1])
{
res++;
intervals[i][1]=intervals[i-1][1];
}
}
return res++;
方法二:先找出交叉區間
static bool cmp(vector<int>& a, vector<int>& b)
{
return a[1]<b[1];
}
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
if(intervals.size()==0) return 0;
sort(intervals.begin(),intervals.end(),cmp);
int res=1;//未交叉數
int start intervals[0][1];
for(vector<int>interval:intervals)
{
if(start<=interval[0])
{
res++;
start=piont[1];
}
}
return intervals.size() - res;
2.引爆氣球 :尋找交叉區間,不過邊界點重疊也算!!
static bool cmp(vector<int>& a, vector<int>& b)
{
return a[1]<b[1];
}
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
if(intervals.size()==0) return 0;
sort(intervals.begin(),intervals.end(),cmp);
int res=1;//未交叉數
int start intervals[0][1];
for(vector<int>interval:intervals)
{
if(start<interval[0]) //邊界重合也算交叉!!
{
res++;
start=interval[1];
}
}
return res; //注意返回值
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