Problem
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.
Example
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
Note
有两种方法。一种是利用Stack去找同一层的两个边,不断累加寄存。如[2, 1, 0, 1, 2],2入栈,1入栈,0入栈,下一个1大于栈顶元素0,则计算此处的雨水量加入res,此过程中0从栈中弹出,1入栈,到下一个2,先弹出1,由于之前还有一个1在栈中,所以计算时高度的因数相减为0,雨水量为0,res无变化,继续pop出栈中的元素,也就是之前的1,此时stack中仍有元素2,说明左边还有边,继续计算底层高度为1,两个值为2的边之间的水量,加入res。
双指针法的思想:先找到左右两边的第一个峰值作为参照位,然后分别向后(向前)每一步增加该位与参照位在这一位的差值,加入sum,直到下一个峰值,再更新为新的参照位。这里有一个需要注意的地方,为什么要先计算左右两个峰值中较小的那个呢?因为在两个指针逼近中间组成最后一个积水区间时,要用较短边计算。
Solution
1. Stack
public class Solution {
public int trapRainWater(int[] A) {
Stack<Integer> stack = new Stack<Integer>();
int res = 0;
for (int i = 0; i < A.length; i++) {
if (stack.isEmpty() || A[i] < A[stack.peek()]) stack.push(i);
else {
while (!stack.isEmpty() && A[i] > A[stack.peek()]) {
int pre = stack.pop();
if (!stack.isEmpty()) {
res += (Math.min(A[i], A[stack.peek()]) - A[pre]) * (i - stack.peek() - 1);
}
}
stack.push(i);
}
}
return res;
}
}
2. Two-Pointer
public class Solution {
public int trap(int[] A) {
int left = 0, right = A.length-1, res = 0;
while (left < right && A[left] < A[left+1]) left++;
while (left < right && A[right] < A[right-1]) right--;
while (left < right) {
int leftmost = A[left], rightmost = A[right];
if (leftmost < rightmost) {
while (left < right && A[++left] < leftmost) res += leftmost - A[left];
}
else {
while (left < right && A[--right] < rightmost) res += rightmost - A[right];
}
}
return res;
}
}
双指针法update: 2018-3
public class Solution {
public int trap(int[] A) {
if (A == null || A.length < 3) return 0;
//set left/right pointers
int l = 0, r = A.length-1;
//find the first left/right peaks as left/right bounds
while (l < r && A[l] <= A[l+1]) l++;
while (l < r && A[r] <= A[r-1]) r--;
//output
int trappedWater = 0;
//implementation
while (l < r) {
int left = A[l];
int right = A[r];
//when you have a higher RIGHT bound...
if (left <= right) {
//when 'l' is highest left bound
//it's safe to add trapped water RIGHTward
while (l < r && left >= A[l+1]) {
l++;
trappedWater += left - A[l];
}
//when left pointer 'l' found 'l+1' a higher left bound
//reset the left bound
l++;
}
//when you have a higher LEFT bound...
else {
//when 'r' is highest right bound
//it's safe to add trapped water LEFTward
while (l < r && right >= A[r-1]) {
r--;
trappedWater += right - A[r];
}
//when right pointer 'r' found 'r-1' a higher right bound
//reset the right bound
r--;
}
}
return trappedWater;
}
}
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