Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
此题和之前的container with most water有些类似，most water里面求的是装最多的水，且里面没有bar，而这道题里面水在哪个bar已经告诉我们了，并且中间由很多的bar，也是用two pointer来做，判定条件也几乎一样，看左右两边哪个bar小哪个进行移动，这里面需要设定两个maxleft，maxright变量，分别表示指针走过的额区域里面bar最大的高度，如果指针所到之处没有max大，则max-pointer为该指针所处位置的水深，否则（大于等于）max为指针所处bar的高度，代码如下：
public class Solution {
public int trap(int[] height) {
int left = 0;
int right = height.length-1;
int maxleft = 0;
int maxright = 0;
int sum = 0;
while(left<right){
if(height[left]<=height[right]){
if(height[left]>=maxleft){
maxleft = height[left];
}else{
sum+=maxleft-height[left];
}
left++;
}else{
if(height[right]>=maxright){
maxright = height[right];
}else{
sum+=maxright-height[right];
}
right--;
}
}
return sum;
}
}