最大面积
题意是给你 a1, a2, …, an 这 n 个数，代表 (i, ai) 坐标，让你从中找两个点与 x 轴围成的容器可以容纳最多的水。
public int maxArea(int[] height)
{
int left = 0,right = height.length - 1;
int max = 0, h = 0;
while(left < right)
{
h = Math.min(height[left],height[right]);
max = Math.max(max, (right-left)*h);
while(height[left]<= h && left < right)
left++;
while(height[right] <= h && left <right)
right--;
}
return max;
}
合并区间
Description
Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considerred overlapping.
Tags: Array, Sort
题意是给你一组区间，让你把区间合并成没有交集的一组区间。我们可以把区间按 start 进行排序，然后遍历排序后的区间，如果当前的 start 小于前者的 end，那么说明这两个存在交集，我们取两者中较大的 end 即可；否则的话直接插入到结果序列中即可。
class Interval
{
int start;
int end;
Interval() { start = 0; end = 0; }
Interval(int s, int e) { start = s; end = e; }
}
class Solution
{
public static void printIntervalList(List<Interval> list)
{
for (Interval interval :list)
{
System.out.print("["+interval.start+","+interval.end+"]"+" ");
}
System.out.println();
}
public static void main(String[] args)
{
List<Interval> intervals = new ArrayList<>();
intervals.add(new Interval(1,3));
intervals.add(new Interval(2,6));
intervals.add(new Interval(8,10));
intervals.add(new Interval(15,18));
printIntervalList(new Solution().merge(intervals));
}
public List<Interval> merge(List<Interval> intervals)
{
if(intervals == null || intervals.size()==0)
return intervals;
intervals.sort(new Comparator<Interval>()
{
@Override
public int compare(Interval o1, Interval o2)
{
if (o1.start < o2.start)
return -1;
if (o1.start > o2.start)
return 1;
return 0;
}
});
int start = intervals.get(0).start;
int end = intervals.get(0).end;
List<Interval> ans = new ArrayList<>();
for(Interval interval : intervals)
{
if(interval.start <= end)
end = Math.max(end, interval.end);
else
{
ans.add(new Interval(start,end));
start = interval.start;
end = interval.end;
}
}
ans.add(new Interval(start,end));
return ans;
}
判断回文数
一半长度
？？
什么时候会走 halfReverseX == x这个return 路径呢？？
？？
好好思考下是否需要计算整个长度，比如 1234321，其实不然，我们只需要计算一半的长度即可，就是在计算过程中的那个逆序数比不断除 10 的数大就结束计算即可，但是这也带来了另一个问题，比如 10 的倍数的数 10010，它也会返回 true，所以我们需要对 10 的倍数的数再加个判断即可，代码如下所示。
public boolean isPalindrome(int x)
{
if (x<0 || (x!= 0 && x%10 == 0))
{
return false;
}
int halfReverseX = 0;
while(x > halfReverseX)
{
halfReverseX = halfReverseX * 10 + x%10;
x /= 10;
}
return halfReverseX == x || halfReverseX / 10 == x;
}
直接求出回文数后进行判断
class Solution
{
public int reverse(int x)
{
long res = 0;
for (;x!=0;x/=10)
{
res = res * 10 + x%10;
}
if(res > Integer.MAX_VALUE || res<Integer.MIN_VALUE)
return 0;
else
return (int) res;
}
public boolean isPalindrome(int x)
{
if(x<0)
return false;
int reverseX = reverse(x);
if (reverseX == x)
return true;
return false;
}
}
整数反转
Description
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
class Solution
{
public static void main(String[] args)
{
System.out.println(new Solution().reverse(-123));
}
public int reverse(int x)
{
long res = 0;
for (;x!=0;x/=10)
{
res = res * 10 + x%10;
}
if(res > Integer.MAX_VALUE || res<Integer.MIN_VALUE)
return 0;
else
return (int) res;
}
// public int reverse(int x)
// {
// long res = 0;
// for( ; x!=0 ; x/=10)
// {
// res = res * 10 + x%10;
// }
//
// if (res>Integer.MAX_VALUE || res < Integer.MIN_VALUE)
// return 0;
// else
// return (int )res;
// }
}
两个大数的加法
以链表表示一个数，低位在前，高位在后，所以题中的例子就是 342 + 465 = 807，所以我们模拟计算即可。
Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
Tags: Linked List, Math
class Solution
{
public static void printList(ListNode list)
{
ListNode current = list;
while(current != null)
{
System.out.print(current.val + " ");
current = current.next;
}
System.out.println();
}
public static void main(String[] args)
{
ListNode l1 = new ListNode(2);
l1.next = new ListNode(4);
l1.next.next = new ListNode(3);
ListNode l2 = new ListNode(5);
l2.next = new ListNode(6);
l2.next.next = new ListNode(4);
printList(new Solution().addTwoNumbers(l1, l2));
}
public ListNode addTwoNumbers (ListNode list1,ListNode list2)
{
ListNode newList = new ListNode(0);
ListNode currentNode = newList;
ListNode node1 = list1,node2 = list2;
int sum =0 ;
while(node1 != null || node2 != null)
{
sum = sum / 10;
if(node1!= null)
{
sum += node1.val;
node1 = node1.next;
}
if( node2 != null)
{
sum += node2.val;
node2 = node2.next;
}
currentNode.next = new ListNode( sum% 10);
currentNode = currentNode.next;
}
if (sum / 10 > 0)
currentNode.next = new ListNode(1);
return newList.next;
}