【LeetCode】Trapping Rain Water解題報告

標籤：trapping rain water   two pointer   one pass   

【題目】

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.

【解析】

題意：求數組組成的凹槽能盛水的容積。

思路：從兩端向中間靠攏，求以兩端為邊的容器能盛水的容積，然後所有值都減去短邊的值，找到新的兩邊，不斷迭代下去直到中間。

public class Solution {    public int trap(int[] A) {        if (A.length < 3) return 0;                int ans = 0;        int l = 0, r = A.length - 1;        while (l < r) {            // find the left edge and right edge            while (l < r && A[l] == 0) l++;            while (l < r && A[r] == 0) r--;                        //get the shoter edge, which decides the volum            int min = Math.min(A[l], A[r]);                         // compute the volum from A[l] to A[r]            int tmp = 0;            for (int i = l; i <= r; i++) {                if (A[i] >= min) {                    A[i] -= min;                } else {                    tmp += min - A[i];                    A[i] = 0;                }            }                        // add to the result            ans += tmp;        }        return ans;    }}

【升級版】【O(n)】

首先找到有效兩邊，如 [0, 1, 2, 3, 0, 3, 2, 1, 0]，遞增的左邊和遞減的右邊是不能盛水的。

然後選擇較短的邊，如果是左邊就開始右移，直到找到一個比左邊高的邊，更新左邊；如果是右邊較短，就左移，直到找到一個更高的邊，更新右邊。

public class Solution {    public int trap(int[] A) {        if (A.length < 3) return 0;                int ans = 0;        int l = 0, r = A.length - 1;                // find the left and right edge which can hold water        while (l < r && A[l] <= A[l + 1]) l++;        while (l < r && A[r] <= A[r - 1]) r--;                while (l < r) {            int left = A[l];            int right = A[r];            if (left <= right) {                // add volum until an edge larger than the left edge                while (l < r && left >= A[++l]) {                    ans += left - A[l];                }            } else {                // add volum until an edge larger than the right volum                while (l < r && A[--r] <= right) {                    ans += right - A[r];                }            }        }        return ans;    }}

【LeetCode】Trapping Rain Water解題報告