算法训练day38 LeetCode435.763.56.
435.无重叠区间
题目
题解
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首先按左边界排列范围
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再将长的重叠区间去除---并记录去除个数
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class Solution { public: static bool cmp(const vector<int> &a, const vector<int> &b) { return a[0] < b[0]; } int eraseOverlapIntervals(vector<vector<int>> &intervals) { if (intervals.size() == 0) return 0; sort(intervals.begin(), intervals.end(), cmp); //将范围按左边界从小到大排序 int count = 0; int end = intervals[0][1]; for (int i = 1; i < intervals.size(); i++) { if (intervals[i][0] >= end) end = intervals[i][1]; else { end = min(end, intervals[i][1]); //可以认为是去除了范围更大的区间 count++; } } return count; } };
763.划分字母区间
题目
题解
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- 重叠区间的思路,统计字符出现的始末位置,排序,找到互不重叠的分组
- 从头遍历字符,并更新字符的最远出现下标,如果找到字符最远出现位置下标和当前下标相等了,则找到了分割点
class Solution { public: vector<int> partitionLabels(string s) { int hash[27] = {0}; for (int i = 0; i < s.size(); i++) { hash[s[i] - 'a'] = i; } vector<int> result; int left = 0; int right = 0; for (int i = 0; i < s.size(); i++) { right = max(right, hash[s[i] - 'a']); //寻找边界 if (i == right) { result.push_back(right - left + 1); left = i + 1; } } return result; } };
56. 合并区间
题目
题解
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class Solution { public: vector<vector<int>> merge(vector<vector<int>> &intervals) { vector<vector<int>> result; if (intervals.size() == 0) return result; sort(intervals.begin(), intervals.end(), [](const vector<int> &a, const vector<int> &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]) // 发现重叠区间 { result.back()[1] = max(result.back()[1], intervals[i][1]); } else { result.push_back(intervals[i]); // 区间不重叠 } } return result; } };