🎉 initial commit for leetcode 763 using find_last_of

Author: liuchuo

C++/763. Partition Labels.cpp

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+763. Partition Labels
+A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.
+
+Example 1:
+Input: S = "ababcbacadefegdehijhklij"
+Output: [9,7,8]
+Explanation:
+The partition is "ababcbaca", "defegde", "hijhklij".
+This is a partition so that each letter appears in at most one part.
+A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
+Note:
+
+S will have length in range [1, 500].
+S will consist of lowercase letters ('a' to 'z') only.
+
+题目大意：给出一个小写字母的字符串S. 我们想把这个字符串分成尽可能多的部分，这样每个字母最多只出现一个部分，并返回一个表示这些部分大小的整数列表。
+分析：遍历字符串，找到S[i]在S中最后一次出现的位置标记为end，end位置是当前要切割的字串部分最短的结尾处，当i == end时候说明start～end可以组成一个最短部分串，将这个部分串的长度(end - start + 1)放入ans数组中，然后将start标记为下一个部分串的开始(end+1)，最后返回ans数组～
+
+class Solution {
+public:
+    vector<int> partitionLabels(string S) {
+        vector<int> ans;
+        for (int i = 0, start = 0, end = 0; i < S.length(); i++) {
+            end = max(end, (int)S.find_last_of(S[i]));
+            if (i == end) {
+                ans.push_back(end - start + 1);
+                start = end + 1;
+            }
+        }
+        return ans;
+    }
+};