解法:判斷兩個list.val誰比較小,就持續走訪誰,並將較小的加入到新的鏈結中
JavaScript
function ListNode(val, next) {
this.val = (val === undefined ? 0 : val)
this.next = (next === undefined ? null : next)
}
/**
* @param {ListNode} list1
* @param {ListNode} list2
* @return {ListNode}
*/
var mergeTwoLists = function (list1, list2) {
let headNode = new ListNode();
let nextNode = headNode;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
nextNode.next = list1;
list1 = list1.next;
} else {
nextNode.next = list2;
list2 = list2.next;
}
nextNode = nextNode.next;
}
nextNode.next = list1 || list2;
return headNode.next;
};