Trapping Rain Water LeetCode JavaScript JS_var trap = function(height) { };-优快云博客

本文介绍了一种计算给定地形图中能存储多少雨水的方法。通过分析地形高度数据，利用两次遍历策略找到最大存储量：首先从左至右确定左侧最高点；接着从右至左确定右侧最高点。

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!

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开始想的是从左边开始扫到右边，取A[i]左边的最大值和右边的最大值中的最小值再减去A[i]的值就是当前格的存水量

var trap = function(height) {
    if(height.length<2)
    {
        return 0;
    }
    var result=0;
    for(var i = 1 ; i < height.length-1;i++){
        var arrayLeft=height.slice(0,i);
        var arrayRight=height.slice(i,height.length);
        var leftMax=Math.max.apply( Math, arrayLeft );
        var rightMax=Math.max.apply(Math, arrayRight );
        var min=Math.min(leftMax,rightMax);
        if(min>height[i]){
            result+=(min-height[i]);
        }
    }
    return result;
};

哦超时了。。

那就优化一下，不要每次都遍历取最大值。

var trap = function(height) {
    if(height.length<2)
    {
        return 0;
    }
    var result=0;
    var leftMax=0;
    var rightMax=Math.max.apply(Math,height.slice(2,height.length));
    for(var i = 1 ; i < height.length-1;i++){
        var arrayLeft=height.slice(0,i);
        var arrayRight=height.slice(i,height.length);
        leftMax=Math.max(height[i-1],leftMax);
        if(rightMax==height[i]){
            rightMax=Math.max.apply(Math,height.slice(i+1,height.length));
        }
        var min = Math.min(rightMax,leftMax);
        if(min>height[i]){
            result+=(min-height[i]);
        }
    }

    return result;
};

哦还是超时了。百度了一下，思路是先看最高点，然后从两边遍历到中间。

var trap = function(height) {
    if(height.length<2)
    {
        return 0;
    }
    var result=0;
    var maxPosition=height.indexOf(Math.max.apply(Math,height));
    var leftMax=height[0];
    for(var i = 1; i < maxPosition ; i++){
        if(leftMax>height[i]){
            result+=leftMax-height[i];
        }
        else{
            leftMax=height[i];
        }
    }
    var rightMax=height[height.length-1];
    for(var j=height.length-2;j>maxPosition;j--){
        if(rightMax>height[j]){
            result+=rightMax-height[j];
        }
        else{
            rightMax=height[j];
        }
    }

    return result;
};

AC啦哈哈哈