| 5|[Longest Palindromic Substring][005]| String|

| 6|[ZigZag Conversion][006]| String|

| 8|[String to Integer (atoi)][008]| Math, String|

| 11|[Container With Most Water][011]| Array, Two Pointers|

| 15|[3Sum][015]| Array, Two Pointers|

| 17|[Letter Combinations of a Phone Number][017]| String, Backtracking|

| 19|[Remove Nth Node From End of List][019]| Linked List, Two Pointers|

[005]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/005/README.md

[006]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/006/README.md

[008]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/008/README.md

[011]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/011/README.md

[015]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/015/README.md

[017]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/017/README.md

[019]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md

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。

不明白的话可以看数据为`1 8 6 2 5 4 8 3 7` 所示的图。

如果用暴力法求每种情况的结果，其时间复杂度为 O(n^2)，相信肯定会超时，我们可以探索下是否有更巧妙的办法呢，题目的标签有双指针，是否就可以想到首尾各放一指针，然后根据条件来收缩。首先计算一次首尾构成的最大面积，然后分析下该移动哪个指针，如果移动大的那个指针的话，那样只会减小面积，所以我们要移动小的那个指针，小的那个指针移动到哪呢？当然是移动到大于之前的值的地方，否则面积不都比之前小么，然后继续更新最大值即可，借助如上分析写出如下代码应该不是什么难事了吧。

classSolution {

publicintmaxArea(int[]height) {

int l=0, r= height.length-1;

int max=0, h=0;

while (l< r) {

h=Math.min(height[l], height[r]);

max=Math.max(max, (r- l)* h);

while (height[l]<= h&& l< r)++l;

while (height[r]<= h&& l< r)--r;

}

return max;

}

}

packagecom.blankj.medium._011;

publicclassSolution {

publicintmaxArea(int[]height) {

intl =0,r =height.length -1;

intmax =0,h =0;

while (l <r) {

h =Math.min(height[l],height[r]);

max =Math.max(max, (r -l) *h);

while (height[l] <=h &&l <r) ++l;

while (height[r] <=h &&l <r) --r;

}

returnmax;

}

publicstaticvoidmain(String[]args) {

Solutionsolution =newSolution();

System.out.println(solution.maxArea(newint[]{1,2,4,3}));// 4

System.out.println(solution.maxArea(newint[]{1,8,6,2,5,4,8,3,7}));// 49

}

}