链表题--02----两数相加

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• 两数相加
• • 题目:
• 方法1
• • 分析 : 分成3个步骤
  • 进位问题
  • 代码:
• 方法2
• • 分析:
  • • • 将两个链表看成是相同长度的进行遍历，==如果一个链表较短则在前面补 0==，比如 987 + 23 = 987 + 023 = 1010
  • 代码

https://leetcode.com/problems/add-two-numbers/

1. 长链表L有,短链表s有
2. 长链表L有,短链表s无
3. 长链表L无,短链表s无

    int carry = 0;
    while(l1 != null || l2 != null) {     
            int sum = x + y + carry;            
            carry = sum / 10;
            sum = sum % 10;          
    }

public class Code05_AddTwoNumbers {

	// 不要提交这个类
	public static class ListNode {
		public int val;
		public ListNode next;

		public ListNode(int val) {
			this.val = val;
		}

		public ListNode(int val, ListNode next) {
			this.val = val;
			this.next = next;
		}
	}

	public static ListNode addTwoNumbers(ListNode head1, ListNode head2) {
		int len1 = listLength(head1);
		int len2 = listLength(head2);
		ListNode l = len1 >= len2 ? head1 : head2;
		ListNode s = l == head1 ? head2 : head1;
		ListNode curL = l;
		ListNode curS = s;
		ListNode last = curL;
		int carry = 0;
		int curNum = 0;
		while (curS != null) {
			curNum = curL.val + curS.val + carry;
			curL.val = (curNum % 10);
			carry = curNum / 10;
			last = curL;
			curL = curL.next;
			curS = curS.next;
		}
		while (curL != null) {
			curNum = curL.val + carry;
			curL.val = (curNum % 10);
			carry = curNum / 10;
			last = curL;
			curL = curL.next;
		}
		if (carry != 0) {
			last.next = new ListNode(1);
		}
		return l;
	}

	// 求链表长度
	public static int listLength(ListNode head) {
		int len = 0;
		while (head != null) {
			len++;
			head = head.next;
		}
		return len;
	}

}

将两个链表看成是相同长度的进行遍历，如果一个链表较短则在前面补 0，比如 987 + 23 = 987 + 023 = 1010

  public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode pre = new ListNode(0);
        ListNode cur = pre;
        int carry = 0;
        while(l1 != null || l2 != null) {
            int x = l1 == null ? 0 : l1.val;
            int y = l2 == null ? 0 : l2.val;
            int sum = x + y + carry;
            
            carry = sum / 10;
            sum = sum % 10;
            cur.next = new ListNode(sum);

            cur = cur.next;
            if(l1 != null)
                l1 = l1.next;
            if(l2 != null)
                l2 = l2.next;
        }
        if(carry == 1) {
            cur.next = new ListNode(carry);
        }
        return pre.next;
    }