第一周：[leetcode] 42. Trapping Rain Water

题目链接：链接

question： 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.

Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.



简而言之，该题是计算非负数组形成的凹槽所能承载的水量；

通过观察得知，数组的载水量由两值形成的局部（凹槽）的最低点决定，因此，计算总在水量应是从数组局部较低点向较高点扫描。将整个数组视为整体，维护left=0和right=height.size()-1两个index，通过比较取较小值向高点移动，直到left == right停止，计算累计在水量。

c++实现如下：

int trap(vector<int>& height) {
int front = 0;
int back = height.size()-1;
int res = 0;
while(front < back){
//取两者较小值向高点移动
if(height[front] < height[back]){
int f = height[front];
while(height[front] <= f && front < back){
res += f - height[front];
front++;
}
}
else{
int b = height[back];
while(height[back] <= b && front < back){
res += b - height[back];
back--;
}
}
}
return res;
}

通过学习solution，发现更加简便的计算方法：

维护level，从第0层开始由两边向中扫描，累积在水量。直至l

int trap(vector<int>& height) {
int l = 0, r = height.size()-1, level = 0, water = 0;
while (l < r) {
int lower = height[height[l] < height[r] ? l++ : r--];
level = max(level, lower);
water += level - lower;
}
return water;
}