LC 763. Partition Labels - Weekly Contest 67 - LeetCode

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.


这题要求最多的partition。
那么我们从左到右遍历一遍，用贪心算法。
能分马上分。除非这个字母在后边还出现而不可分离。
那么聪明的读者可能想到了，我怎么知道这个/些字符在后边还有没有，有的话在多后？
我这里采用的是先扫一遍用HashMap把每个char的起始坐标给缓存起来，
第二遍直接查缓存，进而得出结果。

时间复杂度就是O(n)

	class Solution {
	/**
	* @param S
	* @return
	*/
	public List<Integer> partitionLabels(String S) {
	if (S == null) {
	return new LinkedList<>();
	}
	// core logic
	HashMap<Character, Pair> lookup = new HashMap<>();
	for (int i = 0; i < S.length(); i++) {
	if (lookup.containsKey(S.charAt(i))) {
	lookup.get(S.charAt(i)).end = i;
	} else {
	lookup.put(S.charAt(i), new Pair(i, i));
	}
	}
	LinkedList<Integer> ends = new LinkedList<>();
	for (int i = 0; i < S.length(); ) {
	int end = Math.max(i, lookup.get(S.charAt(i)).end);
	while (i < end) {
	i++;
	end = Math.max(end, lookup.get(S.charAt(i)).end);
	}
	ends.add(end + 1);
	i++;
	}
	LinkedList<Integer> results = new LinkedList<>();
	results.add(ends.get(0));
	for (int i = 1; i < ends.size(); i++) {
	results.add(ends.get(i) - ends.get(i - 1));
	}
	return results;
	}
	private class Pair {
	int start;
	int end;
	public Pair(int start, int end) {
	this.start = start;
	this.end = end;
	}
	@Override
	public String toString() {
	return "Pair{" +
	"start=" + start +
	", end=" + end +
	'}';
	}
	}
	}

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