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Commit5747297

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feat: add 015
1 parentaafe9d1 commit5747297

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4 files changed

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4 files changed

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‎README.md

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@@ -60,6 +60,7 @@
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|2|[Add Two Numbers][002]|Linked List, Math|
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|3|[Longest Substring Without Repeating Characters][003]|Hash Table, Two Pointers, String|
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|8|[String to Integer (atoi)][008]|Math, String|
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|15|[3Sum][015]|Array, Two Pointers|
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|19|[Remove Nth Node From End of List][019]|Linked List, Two Pointers|
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|554|[Brick Wall][554]|Hash Table|
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[002]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/002/README.md
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[003]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/003/README.md
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[008]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/008/README.md
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[015]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/015/README.md
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[019]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md
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[554]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/554/README.md
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‎note/010/README.md

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@@ -30,6 +30,38 @@ isMatch("aab", "c*a*b") → true
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题意是
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```java
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classSolution {
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publicbooleanisMatch(Strings,Stringp) {
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if (p.length()==0)return s.length()==0;
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if (p.length()==1) {
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if (s.length()<1)returnfalse;
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if (p.charAt(0)!= s.charAt(0)&& p.charAt(0)!='.')returnfalse;
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returntrue;
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}
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if (p.charAt(1)!='*') {
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if (s.length()<1)returnfalse;
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if (p.charAt(0)!= s.charAt(0)&& p.charAt(0)!='.')returnfalse;
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return isMatch(s.substring(1), p.substring(1));
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}else {
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// match 0 preceding element
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if (isMatch(s, p.substring(2)))returntrue;
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int i=0;
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// match 1 or more preceding element
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while (i< s.length()&& (p.charAt(0)== s.charAt(i)|| p.charAt(0)=='.')) {
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if (isMatch(s.substring(i+1), p.substring(2))) {
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returntrue;
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}
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++i;
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returnfalse;
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}
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}
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}
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}
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```
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```java
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classSolution {
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publicbooleanisMatch(Strings,Stringp) {

‎note/015/README.md

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#[3Sum][title]
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##Description
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Given an array*S* of*n* integers, are there elements*a*,*b*,*c* in*S* such that*a* +*b* +*c* = 0? Find all unique triplets in the array which gives the sum of zero.
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**Note:** The solution set must not contain duplicate triplets.
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```
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For example, given array S = [-1, 0, 1, 2, -1, -4],
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A solution set is:
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[
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[-1, 0, 1],
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[-1, -1, 2]
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]
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```
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**Tags:** Array, Two Pointers
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##思路
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题意是让你从数组中找出所有三个和为0的元素构成的非重复序列，这样的话我们可以把数组先做下排序，然后遍历这个排序数组，用两个指针分别指向当前元素的下一个和数组尾部，判断三者和与0的大小来移动两个指针，其中细节操作就是注意去重。
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```java
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classSolution {
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publicList<List<Integer>>threeSum(int[]nums) {
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Arrays.sort(nums);
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List<List<Integer>> list=newArrayList<>();
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int i=0;
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while (i< nums.length-2) {
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if (nums[i]>0)break;
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int left= i+1, right= nums.length-1;
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while (left< right) {
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int sum= nums[i]+ nums[left]+ nums[right];
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if (sum==0) {
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list.add(Arrays.asList(nums[i], nums[left++], nums[right--]));
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while (left< right&& nums[left]== nums[left-1])++left;
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while (left< right&& nums[right]== nums[right+1])--right;
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}elseif (sum<0)++left;
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else--right;
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}
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while (nums[i]== nums[++i]&& i< nums.length-2) ;
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}
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return list;
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}
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}
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```
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##结语
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如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
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[title]:https://leetcode.com/problems/3sum
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[ajl]:https://github.com/Blankj/awesome-java-leetcode
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packagecom.blankj.medium._015;
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importjava.util.ArrayList;
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importjava.util.Arrays;
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importjava.util.List;
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/**
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* <pre>
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* author: Blankj
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* blog : http://blankj.com
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* time : 2017/04/23
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* desc :
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* </pre>
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*/
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publicclassSolution {
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publicList<List<Integer>>threeSum(int[]nums) {
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Arrays.sort(nums);
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List<List<Integer>>list =newArrayList<>();
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inti =0;
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while (i <nums.length -2) {
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if (nums[i] >0)break;
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intleft =i +1,right =nums.length -1;
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while (left <right) {
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intsum =nums[i] +nums[left] +nums[right];
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if (sum ==0) {
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list.add(Arrays.asList(nums[i],nums[left++],nums[right--]));
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while (left <right &&nums[left] ==nums[left -1]) ++left;
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while (left <right &&nums[right] ==nums[right +1]) --right;
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}elseif (sum <0) ++left;
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else --right;
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}
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while (nums[i] ==nums[++i] &&i <nums.length -2) ;
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}
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returnlist;
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}
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publicstaticvoidmain(String[]args) {
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Solutionsolution =newSolution();
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System.out.println(solution.threeSum(newint[]{-1,0,1,2, -1, -4}));
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}
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}

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