代码随想录算法训练营第三十六天| LeetCode435. 无重叠区间 763.划分字母区间 56. 合并区间

435. 无重叠区间

题目：力扣

class Solution {
public:
    static const bool cmp(vector& a,vector& b ){
        if(a[1] == b[1]) return a[0] < b[0];
        return a[1] < b[1];
    }
    int eraseOverlapIntervals(vector>& intervals) {
        sort(intervals.begin(),intervals.end(),cmp);
        int ans = 0;
        int end = 0;
        int start = 0;
        int min_end = INT_MAX;
        for(int i = 0; i < intervals.size(); ++i){
            start = intervals[i][0];
            end = intervals[i][1];
            min_end = min(end,min_end);
            if(start >= min_end){
                ans++;
                min_end = end;
            }
        }
        ans = intervals.size() - ans - 1;
        return ans;
    }
};

763.划分字母区间

题目：力扣

//转化为合并区间来做
class Solution {
public:
    static const bool cmp(vector& a,vector& b){
        if(a[0] == b[0]) return a[1] < b[1];
        return a[0] < b[0];
    }
    vector partitionLabels(string s) {
        unordered_map> num;
        vector ans;
        for(int i = 0; i < s.size(); ++i){
            if(num.find(s[i]) == num.end()){
                num[s[i]]= {i,i};
                
            }else{
                num[s[i]][1] = i;
            }
        }
        vector > mind;
        for(auto a : num){
            mind.push_back(a.second);
        }
        sort(mind.begin(),mind.end(),cmp);
        mind.push_back({INT_MAX,INT_MAX});
        int start = 0;
        int end = 0;
        int max_end = -1;
        int min_start = INT_MAX;
        for(int i = 0; i < mind.size(); ++i){
            start = mind[i][0];
            end = mind[i][1];
            min_start = min(start,min_start);
            if(max_end != -1 && start >= max_end){

                ans.push_back(max_end - min_start + 1);
                min_start = start;
            }
            max_end = max(end,max_end);
        }
        return ans; 
    }
};


//另一种方法，统计字符出现的最后位置，然后遍历，如果最远边界和下标相等则记录
class Solution {
public:
    vector partitionLabels(string S) {
        int hash[27] = {0}; // i为字符，hash[i]为字符出现的最后位置
        for (int i = 0; i < S.size(); i++) { // 统计每一个字符最后出现的位置
            hash[S[i] - 'a'] = i;
        }
        vector result;
        int left = 0;
        int right = 0;
        for (int i = 0; i < S.size(); i++) {
            right = max(right, hash[S[i] - 'a']); // 找到字符出现的最远边界
            if (i == right) {
                result.push_back(right - left + 1);
                left = i + 1;
            }
        }
        return result;
    }
};

56. 合并区间

题目：力扣

class Solution {
public:
    static const bool cmp(vector& a, vector& b){
        return a[0] < b[0];
    } 
    vector> merge(vector>& intervals) {
        sort(intervals.begin(),intervals.end(),cmp);
        vector> ans;
        intervals.push_back({INT_MAX,INT_MAX});
        int start = 0;
        int end = 0;
        int min_start = INT_MAX;
        int max_end = -1;
        for(int i = 0; i < intervals.size(); ++i){
            start = intervals[i][0];
            end = intervals[i][1];
            min_start = min(start,min_start);
            if(max_end != -1 && start > max_end){
                ans.push_back({min_start,max_end});
                min_start = start;
            }
            max_end = max(end,max_end);
        }
        return ans;
    }
};


//另一种方法，不断合并大的区间即可
class Solution {
public:
    vector> merge(vector>& intervals) {
        vector> result;
        if (intervals.size() == 0) return result;
        // 排序的参数使用了lambda表达式
        sort(intervals.begin(), intervals.end(), [](const vector& a, const vector& b){return a[0] < b[0];});

        result.push_back(intervals[0]);
        for (int i = 1; i < intervals.size(); i++) {
            if (result.back()[1] >= intervals[i][0]) { // 合并区间
                result.back()[1] = max(result.back()[1], intervals[i][1]);
            } else {
                result.push_back(intervals[i]);
            }
        }
        return result;
    }
};

总结

题型：贪心（重叠区间）

技巧：排序+遍历。