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Commit2121604

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[LEET-1111] add more problem descriptions
1 parente8cb699 commit2121604

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3 files changed

+51
-26
lines changed

3 files changed

+51
-26
lines changed

‎leetcode-algorithms/src/main/java/com/stevesun/solutions/Flatten2DVector.java

Lines changed: 47 additions & 23 deletions
Original file line numberDiff line numberDiff line change
@@ -5,34 +5,58 @@
55
importjava.util.List;
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importjava.util.Queue;
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/**Implement an iterator to flatten a 2d vector.
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For example,
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Given 2d vector =
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[
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[1,2],
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[3],
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[4,5,6]
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]
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By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,2,3,4,5,6].
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Hint:
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How many variables do you need to keep track?
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Two variables is all you need. Try with x and y.
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Beware of empty rows. It could be the first few rows.
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To write correct code, think about the invariant to maintain. What is it?
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The invariant is x and y must always point to a valid point in the 2d vector. Should you maintain your invariant ahead of time or right when you need it?
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Not sure? Think about how you would implement hasNext(). Which is more complex?
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Common logic in two different places should be refactored into a common method.
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Follow up:
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As an added challenge, try to code it using only iterators in C++ or iterators in Java.*/
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publicclassFlatten2DVector {
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}
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classVector2DimplementsIterator<Integer> {
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privateQueue<Integer>cache;
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privateList<List<Integer>>vec2d;
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publicVector2D(List<List<Integer>>vec2d) {
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this.vec2d =vec2d;
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this.cache =newLinkedList<Integer>();
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if(vec2d !=null &&vec2d.size() >0){
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for(List<Integer>list :vec2d){
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for(inti :list){
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cache.offer(i);
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classVector2DimplementsIterator<Integer> {
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privateQueue<Integer>cache;
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privateList<List<Integer>>vec2d;
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publicVector2D(List<List<Integer>>vec2d) {
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this.vec2d =vec2d;
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this.cache =newLinkedList<Integer>();
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if (vec2d !=null &&vec2d.size() >0) {
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for (List<Integer>list :vec2d) {
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for (inti :list) {
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cache.offer(i);
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}
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}
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}
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}
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}
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@Override
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publicIntegernext() {
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returncache.poll();
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}
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@Override
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publicbooleanhasNext() {
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return !cache.isEmpty();
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@Override
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publicIntegernext() {
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returncache.poll();
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}
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@Override
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publicbooleanhasNext() {
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return !cache.isEmpty();
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}
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}
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}
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}

‎leetcode-algorithms/src/main/java/com/stevesun/solutions/InorderSuccessorInBST.java

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -12,16 +12,16 @@ public class InorderSuccessorInBST {
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//Amazed at this solution: https://discuss.leetcode.com/topic/25698/java-python-solution-o-h-time-and-o-1-space-iterative
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//I used brute force: traverse the tree and store and nodes in a list, then traverse the list to compare with p to return the successor if there is.
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publicTreeNodeinorderSuccessor(TreeNoderoot,TreeNodep) {
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TreeNodesucc =null;
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TreeNodesuccessor =null;
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while(root !=null){
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if(p.val <root.val){
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succ =root;
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successor =root;
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root =root.left;
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}else {
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root =root.right;
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}
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}
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returnsucc;
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returnsuccessor;
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}
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}

‎leetcode-algorithms/src/main/java/com/stevesun/solutions/OneEditDistance.java

Lines changed: 1 addition & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
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packagecom.stevesun.solutions;
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3+
/**Given two strings S and T, determine if they are both one edit distance apart.*/
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publicclassOneEditDistance {
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publicstaticbooleanisOneEditDistance(Strings,Stringt) {

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