代码随想录刷题第36天

今天的题目都与重叠区间有关。第一题是无重叠区间https://leetcode.cn/problems/non-overlapping-intervals/description/，与昨天用箭射气球的逻辑相同，按左边界排序，找出重叠区间数量即可。

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

第二题是划分字母区间https://leetcode.cn/problems/partition-labels/submissions/503106369/，该题是哈希与贪心的综合，首先用哈希方法统计字符串中个字母的最远出现位置，在向后遍历过程中不断记录不同字母的最远位置，到达最远位置时记录结果，更新区间左端位置，再次向后遍历。

class Solution {
public:
    vector partitionLabels(string s) {
        int hash[26] = {0};
        for (int i = 0; i < s.size(); i++){
            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']);
            if (right == i){
                result.push_back(right - left + 1);
                left = right + 1;
            }
        }
        return result;

    }
};

第三题是合并区间https://leetcode.cn/problems/merge-intervals/description/，先判断是否重叠，若重叠，更新右边界，若不重叠，直接讲区间加入结果集中。

class Solution {
public:
static bool cmp(const vector& a, const vector& b){
    return a[0] < b[0];
}
    vector> merge(vector>& intervals) {
        if (intervals.size() == 0) return intervals;
        sort(intervals.begin(), intervals.end(), cmp);
        vector> result;
        result.push_back(intervals[0]);
        for (int i = 1; i < intervals.size(); i++){
            if (result.back()[1] >= intervals[i][0]){//重叠
               result.back()[1] = max(result.back()[1], intervals[i][1]);
            }
            else result.push_back(intervals[i]);
        }
        return result;

    }
};

100题目标达成，给自己留个纪念哈哈加油