leetcode-Trapping Rain Water (2014.3.10)
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每一个节点的存水量为左右最大高度较小值和本节点的差值min(maxLeftHeight[i],maxRightHeight[i])-A[i]。用空间换时间,若每个节点计算会超时,用两个数组记录每个节点的左右最高值。另外,vector不初始化会报错。
class Solution {
public:
int trap(int A[], int n) {
// Start typing your Java solution below
// DO NOT write main() function
vector<int> left(n);
vector<int> right(n); //若vector不初始化会报错
int maxHeight=0;
for(int i=0;i<n;i++){
left[i]=maxHeight;
if(A[i]>maxHeight) maxHeight=A[i];
}
maxHeight=0;
for(int j=n-1;j>=0;j--){
right[j]=maxHeight;
if(A[j]>maxHeight) maxHeight=A[j];
}
int sum=0;
for(int i=0;i<n;i++){
if(min(left[i],right[i])-A[i]>0){
sum+=min(left[i],right[i])-A[i];
}
}
return sum;
}
};
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