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Commitfde517c

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applewjg
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Change-Id: Id40fa127b22fda2949a457e6fa7679ee941e0b13
1 parent433387e commitfde517c

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3 files changed

+100
-9
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3 files changed

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‎CombinationSum.java

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/*
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Author: Annie Kim, anniekim.pku@gmail.com
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Date: May 25, 2013
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Problem: Combination Sum
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Difficulty: Easy
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Source: http://leetcode.com/onlinejudge#question_39
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Notes:
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Given a set of candidate numbers (C) and a target number (T), find all unique
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combinations in C where the candidate numbers sums to T.
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The same repeated number may be chosen from C unlimited number of times.
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Note:
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All numbers (including target) will be positive integers.
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Elements in a combination (a1, a2, .. , ak) must be in non-descending order. (ie, a1 <= a2 <= ... <= ak).
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The solution set must not contain duplicate combinations.
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For example, given candidate set 2,3,6,7 and target 7,
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A solution set is:
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[7]
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[2, 2, 3]
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Solution: Sort & Recursion.
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*/
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publicclassSolution {
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publicList<List<Integer>>combinationSum(int[]candidates,inttarget) {
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List<List<Integer>>res =newArrayList<List<Integer>>();
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Arrays.sort(candidates);
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ArrayList<Integer>path =newArrayList<Integer>();
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combinationSumRe(candidates,target,0,path,res);
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returnres;
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}
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voidcombinationSumRe(int[]candidates,inttarget,intstart,ArrayList<Integer>path,List<List<Integer>>res) {
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if (target ==0) {
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ArrayList<Integer>p =newArrayList<Integer>(path);
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res.add(p);
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return;
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}
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for (inti =start;i <candidates.length &&target >=candidates[i]; ++i) {
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path.add(candidates[i]);
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combinationSumRe(candidates,target-candidates[i],i,path,res);
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path.remove(path.size() -1);
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}
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}
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}

‎MergeIntervals.java

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/*
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Author: King, wangjingui@outlook.com
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Date: Dec 20, 2014
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Problem: Merge Intervals
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Difficulty: Medium
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Source: https://oj.leetcode.com/problems/merge-intervals/
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Notes:
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Given a collection of intervals, merge all overlapping intervals.
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For example,
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Given [1,3],[2,6],[8,10],[15,18],
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return [1,6],[8,10],[15,18].
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Solution: 1. Sort in ascending order of 'start'.
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2. Traverse the 'intervals', merge or push...
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*/
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/**
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* Definition for an interval.
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* public class Interval {
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* int start;
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* int end;
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* Interval() { start = 0; end = 0; }
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* Interval(int s, int e) { start = s; end = e; }
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* }
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*/
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publicclassSolution {
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publicList<Interval>merge(List<Interval>intervals) {
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Comparator<Interval>comp =newComparator<Interval>(){
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publicintcompare(Intervala,Intervalb) {
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if(a.start <b.start) {
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return -1;
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}elseif(a.start >b.start){
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return1;
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}else {
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if (a.end <b.end)return -1;
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elseif (a.end >b.end)return1;
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return0;
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}
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}
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};
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ArrayList<Interval>res =newArrayList<Interval>();
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intN =intervals.size();
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if (N <=1)returnintervals;
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Collections.sort(intervals,comp);
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Intervallast =intervals.get(0);
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for (inti =0;i <N; ++i) {
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if (intervals.get(i).start >last.end) {
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res.add(last);
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last =intervals.get(i);
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}else {
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last.end =Math.max(last.end,intervals.get(i).end);
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}
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}
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res.add(last);
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returnres;
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}
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}

‎MergekSortedLists.java

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Solution: Find the smallest list-head first using minimum-heap(lgk).
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complexity: O(N*KlgK)
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*/
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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/**
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* Definition for singly-linked list.
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* public class ListNode {

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