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;
    }
}