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手写跳表删除逻辑 bugfix #2336

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merged 1 commit into from
Mar 23, 2024
Merged

手写跳表删除逻辑 bugfix #2336

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40 changes: 20 additions & 20 deletions docs/database/redis/redis-skiplist.md
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Original file line number Diff line number Diff line change
Expand Up @@ -233,7 +233,7 @@ private int randomLevel() {
/**
* 默认情况下的高度为1，即只有自己一个节点
*/
private int leveCount = 1;
private int levelCount = 1;

/**
* 跳表最底层的节点，即头节点
Expand Down Expand Up @@ -271,9 +271,9 @@ public void add(int value) {
maxOfMinArr[i].forwards[i] = newNode;
}

//如果当前newNode高度大于跳表最高高度则更新leveCount
if (leveCount < level) {
leveCount = level;
//如果当前newNode高度大于跳表最高高度则更新levelCount
if (levelCount < level) {
levelCount = level;
}

}
Expand All @@ -299,7 +299,7 @@ public void add(int value) {
public Node get(int value) {
Node p = h;
//找到小于value的最大值
for (int i = leveCount - 1; i >= 0; i--) {
for (int i = levelCount - 1; i >= 0; i--) {
while (p.forwards[i] != null && p.forwards[i].data < value) {
p = p.forwards[i];
}
Expand Down Expand Up @@ -334,8 +334,8 @@ public Node get(int value) {
public void delete(int value) {
Node p = h;
//找到各级节点小于value的最大值
Node[] updateArr = new Node[leveCount];
for (int i = leveCount - 1; i >= 0; i--) {
Node[] updateArr = new Node[levelCount];
for (int i = levelCount - 1; i >= 0; i--) {
while (p.forwards[i] != null && p.forwards[i].data < value) {
p = p.forwards[i];
}
Expand All @@ -344,16 +344,16 @@ public void delete(int value) {
//查看原始层节点前驱是否等于value，若等于则说明存在要删除的值
if (p.forwards[0] != null && p.forwards[0].data == value) {
//从最高级索引开始查看其前驱是否等于value，若等于则将当前节点指向value节点的后继节点
for (int i = leveCount - 1; i >= 0; i--) {
for (int i = levelCount - 1; i >= 0; i--) {
if (updateArr[i].forwards[i] != null && updateArr[i].forwards[i].data == value) {
updateArr[i].forwards[i] = updateArr[i].forwards[i].forwards[i];
}
}
}

//从最高级开始查看是否有一级索引为空，若为空则层级减1
while (leveCount > 1 && h.forwards[leveCount] == null) {
leveCount--;
while (levelCount > 1 && h.forwards[levelCount - 1] == null) {
levelCount--;
}

}
Expand All @@ -379,7 +379,7 @@ public class SkipList {
/**
* 默认情况下的高度为1，即只有自己一个节点
*/
private int leveCount = 1;
private int levelCount = 1;

/**
* 跳表最底层的节点，即头节点
Expand Down Expand Up @@ -436,9 +436,9 @@ public class SkipList {
maxOfMinArr[i].forwards[i] = newNode;
}

//如果当前newNode高度大于跳表最高高度则更新leveCount
if (leveCount < level) {
leveCount = level;
//如果当前newNode高度大于跳表最高高度则更新levelCount
if (levelCount < level) {
levelCount = level;
}

}
Expand All @@ -463,7 +463,7 @@ public class SkipList {
public Node get(int value) {
Node p = h;
//找到小于value的最大值
for (int i = leveCount - 1; i >= 0; i--) {
for (int i = levelCount - 1; i >= 0; i--) {
while (p.forwards[i] != null && p.forwards[i].data < value) {
p = p.forwards[i];
}
Expand All @@ -484,8 +484,8 @@ public class SkipList {
public void delete(int value) {
Node p = h;
//找到各级节点小于value的最大值
Node[] updateArr = new Node[leveCount];
for (int i = leveCount - 1; i >= 0; i--) {
Node[] updateArr = new Node[levelCount];
for (int i = levelCount - 1; i >= 0; i--) {
while (p.forwards[i] != null && p.forwards[i].data < value) {
p = p.forwards[i];
}
Expand All @@ -494,16 +494,16 @@ public class SkipList {
//查看原始层节点前驱是否等于value，若等于则说明存在要删除的值
if (p.forwards[0] != null && p.forwards[0].data == value) {
//从最高级索引开始查看其前驱是否等于value，若等于则将当前节点指向value节点的后继节点
for (int i = leveCount - 1; i >= 0; i--) {
for (int i = levelCount - 1; i >= 0; i--) {
if (updateArr[i].forwards[i] != null && updateArr[i].forwards[i].data == value) {
updateArr[i].forwards[i] = updateArr[i].forwards[i].forwards[i];
}
}
}

//从最高级开始查看是否有一级索引为空，若为空则层级减1
while (leveCount > 1 && h.forwards[leveCount] == null) {
leveCount--;
while (levelCount > 1 && h.forwards[levelCount - 1] == null) {
levelCount--;
}

}
Expand Down