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优化了0763.划分字母区间.md 新增贪心思路的代码逻辑

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speed
2022-04-07 10:52:27 +08:00
parent f6a1b86c4d
commit 3048b00d72
1 changed files with 18 additions and 10 deletions
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28
problems/0763.划分字母区间.md
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@ -90,35 +90,43 @@ public:
return a[0] < b[0];
}
// 记录每个字母出现的区间
void countLabels(string s, vector<vector<int>> &hash) {
vector<vector<int>> countLabels(string s) {
vector<vector<int>> hash(26, vector<int>(2, INT_MIN));
vector<vector<int>> hash_filter;
for (int i = 0; i < s.size(); ++i) {
if (hash[s[i] - 'a'][0] == INT_MIN) {
hash[s[i] - 'a'][0] = i;
}
hash[s[i] - 'a'][1] = i;
}
// 去除字符串中未出现的字母所占用区间
for (int i = 0; i < hash.size(); ++i) {
if (hash[i][0] != INT_MIN) {
hash_filter.push_back(hash[i]);
}
}
return hash_filter;
}
vector<int> partitionLabels(string s) {
vector<int> res;
vector<vector<int>> hash(26, vector<int>(2, INT_MIN));
countLabels(s, hash);
// 这一步得到的 hash 即为无重叠区间题意中的输入样例格式：区间列表
// 只不过现在我们要求的是区间分割点
vector<vector<int>> hash = countLabels(s);
// 按照左边界从小到大排序
sort(hash.begin(), hash.end(), cmp);
// 记录最大右边界
int rightBoard = INT_MIN;
int rightBoard = hash[0][1];
int leftBoard = 0;
for (int i = 0; i < hash.size(); ++i) {
// 过滤掉字符串中没有的字母
if (hash[i][0] == INT_MIN) {
continue;
}
for (int i = 1; i < hash.size(); ++i) {
// 由于字符串一定能分割，因此,
// 一旦下一区间左边界大于当前右边界，即可认为出现分割点
if (rightBoard != INT_MIN && hash[i][0] > rightBoard) {
if (hash[i][0] > rightBoard) {
res.push_back(rightBoard - leftBoard + 1);
leftBoard = hash[i][0];
}
rightBoard = max(rightBoard, hash[i][1]);
}
// 最右端
res.push_back(rightBoard - leftBoard + 1);
return res;
}