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.
app
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.spa
有兩種方法。一種是利用Stack去找同一層的兩個邊,不斷累加寄存。如[2, 1, 0, 1, 2],2入棧,1入棧,0入棧,下一個1大於棧頂元素0,則計算此處的雨水量加入res,此過程當中0從棧中彈出,1入棧,到下一個2,先彈出1,因爲以前還有一個1在棧中,因此計算時高度的因數相減爲0,雨水量爲0,res無變化,繼續pop出棧中的元素,也就是以前的1,此時stack中仍有元素2,說明左邊還有邊,繼續計算底層高度爲1,兩個值爲2的邊之間的水量,加入res。
雙指針法的思想:先找到左右兩邊的第一個峯值做爲參照位,而後分別向後(向前)每一步增長該位與參照位在這一位的差值,加入sum,直到下一個峯值,再更新爲新的參照位。這裏有一個須要注意的地方,爲何要先計算左右兩個峯值中較小的那個呢?由於在兩個指針逼近中間組成最後一個積水區間時,要用較短邊計算。指針
1. Stackcode
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-Pointerhtm
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; } }
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; } }