[LeetCode]Container With Most Water、Trapping Rain Water
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container.
意思就是选两个数字来组成一个容器，问最多可以得到多少水。其实也就是求最大面积。
面积取决于两个要素：长和高，很明显我们从两头开始向中间逼近的时候，长是在减小的，所以为了面积增加，必须高要增加，对吧。然后两端的高度的较小值跟高度乘积就是当前面积。而两头都可以向中间逼近以获得更大的高度，但每次逼近应当选取较小的那一端。
class Solution {
public:
int maxArea(vector<int> &h) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int n=h.size();
if ( n <=1 )
return 0;
int ans=0;
int l=0,r=n-1;
while(l<r)
{
ans= max(ans,(r-l)*min(h[l],h[r]));
if ( h[l] < h[r] )
{
int preL=h[l];
while(l<r && h[l]<=preL)
l++;
}
else
{
int preR=h[r];
while(l<r&&h[r]<=preR)
r--;
}
}
return ans;
}
};
Trapping Rain Water：
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!
题意比较好理解，如果我们从全局上来考虑，去思考会形成多少个坑，然后计算每个坑能装多少水，就会比较麻烦。简单的思路是：每个坐标位置是否能装水，能装多少水，这个是很容易计算的，最后全部加起来就好了嘛。
那么一个坐标位置能不能装水呢？条件是左右两遍都有比它高的柱子，那么肯定能形成一个坑，然后能装多少水呢？当然是由左右两个高柱子低的那根与自己高度的差值了。配合上面配图，应该比较好理解的。
所以问题转变为，求每个位置，左右最高的高度是多少：
class Solution {
public:
int trap(int A[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
assert(A&&n>=0);
if ( n<=2 )
return 0;
vector<int> lHigh(n,0);
int leftH=0;
for(int i=0;i<n;i++)
{
lHigh[i]=leftH;
leftH=max(leftH,A[i]);
}
int ans=0;
int rightH=0;
for(int i=n-1;i>=0;i--)
{
int h=min(rightH,lHigh[i]);
if ( h>A[i] )
ans+=h-A[i];
rightH=max(rightH,A[i]);
}
return ans;
}
};