題目形貌

刪除有序鏈表中的反覆節點，返轉頭節點

1. 刪撤除一切反覆節點，比方1->1->2->2->3->4，返回3->4
2. 反覆的節點中保存一個，比方1->1->2->2->3->4，返回1->2->3->4

刪撤除一切反覆節點

function ListNode(x){
    this.val = x;
    this.next = null;
}
function deleteDuplication(pHead){
    if(pHead === null || pHead.next === null) 
        return pHead;
    var H = new ListNode(null);
    H.next = pHead;

    var pre = H;
    var cur = pHead;

    while(cur !== null && cur.next !== null) {
        if(cur.next.val === cur.val){
            var curRepetitiveVal = cur.val;
            while(cur !== null && cur.val === curRepetitiveVal) {
                cur = cur.next;
            }
            pre.next = cur;
        }else{
            pre = cur;
            cur = cur.next;
        }
    }

    return H.next;
}

細節
這內里有幾個須要注重的細節：

• 新建一個空的頭節點，由於這內里牽扯到換新的鏈表頭的題目，所以為了輕易新建一個新的節點作為鏈表的頭節點

每一個反覆節點中保存一個

function ListNode(x){
    this.val = x;
    this.next = null;
}
function deleteDuplication(pHead){
    if(pHead === null || pHead.next === null) 
        return pHead;
    var H = new ListNode(null);
    H.next = pHead;

    var pre = H;
    var cur = pHead;

    while(cur !== null && cur.next !== null) {
        if(cur.next.val === cur.val){
            pre = cur;
            var curRepetitiveVal = cur.val;
            while(cur !== null && cur.val === curRepetitiveVal) {
                cur = cur.next;
            }
            pre.next = cur;
        }else{
            pre = cur;
            cur = cur.next;
        }
    }    

    return H.next;
}

細節

• 這個和刪撤除一切反覆節點的區分就是發現有雷同的節點時，pre指針立馬指向反覆的節點中的第一個節點