LeetCode - Trapping Rain Water
Posted 朽木の半夏
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题目描述:
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!
AC代码:
int right(int* a,int n){
int ans = 0;
int has = 0;
int h = 0;
for(int i = 0 ; i < n ; i ++){
if(a[i]>=h){
ans+=has;
h=a[i];
has = 0;
}else{
has+=h-a[i];
}
}
return ans;
}
int left(int*a,int n){
int ans = 0;
int has = 0;
int h = 0;
for(int i = n-1 ; i >= 0; i --){
if(a[i]>h){
ans+=has;
h = a[i];
has = 0;
}else{
has += h-a[i];
}
}
return ans;
}
int trap(int* height, int heightSize) {
return right(height,heightSize)+left(height,heightSize);
} Trapping Rain Water
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