面试中,经常遇到的一个简单算法题:查找两个单链表的公共节点
最近在读react源码的时候发现一个react树中对该算法的运用(见getLowestCommonAncestor函数),在此做简单的记录。
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getParent
在react树中获取当前实例节点的父节点实例
//HostComponent组件对应的DOM,比如App的tag=3, 表示为类组件,其child为tag=5对应div元素。
function getParent(inst) {
do {
inst = inst.return;
// TODO: If this is a HostRoot we might want to bail out.
// That is depending on if we want nested subtrees (layers) to bubble
// events to their parent. We could also go through parentNode on the
// host node but that wouldn't work for React Native and doesn't let us
// do the portal feature.
} while (inst && inst.tag !== HostComponent);
if (inst) {
return inst;
}
return null;
}
getLowestCommonAncestor
获取节点A与B的最近的公共祖先节点
算法题:找到两个链表的公共节点
export function getLowestCommonAncestor(instA, instB) {
//获取子节点A在树中的深度
let depthA = 0;
for (let tempA = instA; tempA; tempA = getParent(tempA)) {
depthA++;
}
//获取子节点B在树中的深度
let depthB = 0;
for (let tempB = instB; tempB; tempB = getParent(tempB)) {
depthB++;
}
// If A is deeper, crawl up.
// 如果A的高度高,那么A节点先往上走depthA - depthB个节点,最后同时走,直到父节点是同一个
while (depthA - depthB > 0) {
instA = getParent(instA);
depthA--;
}
// 如果B的高度高,那么B节点先往上走depthB - depthB个节点,最后同时走,直到父节点是同一个
// If B is deeper, crawl up.
while (depthB - depthA > 0) {
instB = getParent(instB);
depthB--;
}
// Walk in lockstep until we find a match.
// 现在,指针所处的位置的高度一致,可以同时往上查找,直到找到公共的节点
let depth = depthA;
while (depth--) {
if (instA === instB || instA === instB.alternate) {
return instA;
}
instA = getParent(instA);
instB = getParent(instB);
}
return null;
}
isAncestor
判断A节点是否是B节点的祖先节点
export function isAncestor(instA, instB) {
while (instB) {
if (instA === instB || instA === instB.alternate) {
return true;
}
instB = getParent(instB);
}
return false;
}
getParentInstance
对getParent的export封装:
export function getParentInstance(inst) {
return getParent(inst);
}
traverseTwoPhase
对inst及其以上的树执行冒泡捕获的操作,执行fn。类似事件的冒泡捕获
export function traverseTwoPhase(inst, fn, arg) {
const path = [];
//将inst的父节点入栈,数组最后的为最远的祖先
while (inst) {
path.push(inst);
inst = getParent(inst);
}
let i;
//从最远的祖先开始向inst节点捕获执行fn
for (i = path.length; i-- > 0; ) {
fn(path[i], 'captured', arg);
}
//从inst节点开始向最远的祖先节点冒泡执行fn
for (i = 0; i < path.length; i++) {
fn(path[i], 'bubbled', arg);
}
}
traverseEnterLeave
当关注点从from节点移出然后移入to节点的时候,在from执行执行类似移入移出的操作,from节点
export function traverseEnterLeave(from, to, fn, argFrom, argTo) {
const common = from && to ? getLowestCommonAncestor(from, to) : null;
const pathFrom = [];
while (true) {
if (!from) {
break;
}
if (from === common) {
break;
}
const alternate = from.alternate;
if (alternate !== null && alternate === common) {
break;
}
pathFrom.push(from);
from = getParent(from);
}
const pathTo = [];
while (true) {
if (!to) {
break;
}
if (to === common) {
break;
}
const alternate = to.alternate;
if (alternate !== null && alternate === common) {
break;
}
pathTo.push(to);
to = getParent(to);
}
//以上代码将from节点到from与to节点的最近公共祖先节点(不包括公共祖先节点)push到pathFrom数组
//以上代码将to节点到from与to节点的最近公共祖先节点(不包括公共祖先节点)push到pathTo数组
// 以下代码用于对pathFrom冒泡,执行fn
for (let i = 0; i < pathFrom.length; i++) {
fn(pathFrom[i], 'bubbled', argFrom);
}
// 以下代码用于对pathTo捕获,执行fn
for (let i = pathTo.length; i-- > 0; ) {
fn(pathTo[i], 'captured', argTo);
}
}
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