LeetCode 42. Trapping Rain Water

用了6ms,只击败了6%的提交者，表示好坑，之后会进一步改良与完善注释，先就这样：

import java.util.LinkedList;public class Solution {    public static class Pair{        public int height;        public int index;        public Pair(int height, int index) {            super();            this.height = height;            this.index = index;        }    }    public int trap(int[] height) {        if(height==null||height.length<3) return 0;        LinkedList<Pair> list = new LinkedList<>();        int lowest = height[0];        list.add(new Pair(lowest,0));        int sum = 0;        for (int i = 1; i < height.length; i++) {            if(lowest < height[i]){                if(list.size()==1){                    list.pop();//上升阶段                    list.push(new Pair(height[i],i));                    lowest = height[i];                }else{                    while(true){                        if(list.size()==1){                            list.clear();                            list.push(new Pair(height[i],i));                            break;                        }                        Pair behind = list.pop();                        if(behind.height>height[i]){                            list.push(behind);                            break;                        }                        Pair front = list.pop();                        if(front.height>=height[i]){                            sum+=(i-behind.index)*(height[i]-behind.height);                            list.push(front);                            list.push(new Pair(height[i],behind.index));                            break;                        }else{                            sum+=(i-behind.index)*(front.height-behind.height);                            list.push(front);                        }                    }                    lowest = list.peek().height;                }            }            if(lowest > height[i]){//下降阶段                list.push(new Pair(height[i],i));                lowest = height[i];            }        }        return sum ;    }}

看别人做的，基本思路就是记录每个坑需要的注水量，
先把每个‘坑’左侧最高和右侧最高计算出来
需要注水的话 min(左侧最高,右侧最高)-本身高度就是需要的注水量

public class Solution3 {    public int trap(int[] height) {        if(height.length<3){            return 0;        }        int[] maxleft = new int[height.length];        int[] maxRight = new int[height.length];        maxleft[0] = height[0];        for(int i=1;i<height.length;i++){            maxleft[i] = max(maxleft[i-1], height[i]);        }        maxRight[height.length-1] = height[height.length-1];        for(int i=height.length-2;i>0;i--){            maxRight[i] = max(maxRight[i+1], height[i]);        }        int sum = 0;        for (int i = 0; i < height.length; i++) {            sum+=max(0, min(maxleft[i], maxRight[i])-height[i]);        }        return sum;    }    public int max(int a,int b){        return a>b?a:b;    }    public int min(int a,int b){        return a<b?a:b;    }}