Trapping Rain Water LeetCode JavaScript JS

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