题目要求: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!
代码如下:
class Solution {
public:
int trap(int A[], int n) { int maxIdx = 0;
int water = 0; //找到最长的木板,设为maxIdx
for(int i = 1; i < n; i++){
if(A[i] > A[maxIdx]){
maxIdx = i;
}
} int max = A[0]; //左侧逼近
for(int i = 1; i < maxIdx; i++){ if(max < A[i])
max = A[i];
//木板左边(max)和右边(最高)都比它高,则可以放
// max - 该木板长度 的水
else
water += max - A[i];
} //右侧逼近
max = A[n - 1];
for(int i = n - 2; i >= maxIdx; i--){
if(max < A[i]) max = A[i];
else water += max - A[i];
} return water; }
};