LeetCode之Trapping Rain Water

【题目】

Givennnon-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], return6.


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 Marcosfor contributing this image!


【题意】

给定n个非负整数,代表一个柱状图,每一个柱子的宽度为1,计算下雨之后柱状图能装多少水?

例如:

[0,1,0,2,1,0,1,3,2,1,2,1] 返回 6

上述柱状图是由数组表示[0,1,0,2,1,0,1,3,2,1,2,1]。在这种情况下,6个单位的雨水(蓝色部分)被

【分析】

对于每个柱子,找到其左右两边最高的柱子,该柱子能容纳的面积就是 min(leftMostHeight,rightMostHeight) - height。所以,

1. 从左往右扫描一遍,对于每个柱子,求取左边最大值;

2. 从右往左扫描一遍,对于每个柱子,求最大右值;

3. 再扫描一遍,把每个柱子的面积并累加。

也可以,

1. 扫描一遍,找到最高的柱子,这个柱子将数组分为两半;

2. 处理左边一半;

3. 处理右边一半。

【代码】

/*********************************
*   日期:2014-01-20
*   作者:SJF0115
*   题号: Trapping Rain Water
*   来源:http://oj.leetcode.com/problems/trapping-rain-water/
*   结果:AC
*   来源:LeetCode
*   总结:
**********************************/
#include 
#include 
#include 
using namespace std;

class Solution {
public:
    int trap(int A[], int n) {
        if(A == NULL || n < 1)return 0;
        int i;

		int* leftMostHeight = (int*)malloc(sizeof(int)*n);
		int* rightMostHeight = (int*)malloc(sizeof(int)*n);

		int maxHeight = 0;
		for(i = 0; i < n;i++){
			leftMostHeight[i] = maxHeight;
			if(maxHeight < A[i]){
                maxHeight = A[i];
            }
		}

		maxHeight = 0;
		for(i = n-1;i >= 0;i--){
			rightMostHeight[i] = maxHeight;
			if(maxHeight < A[i]){
                maxHeight = A[i];
            }
		}

		int water = 0;
		for(i =0; i < n; i++){
			int curWater = min(leftMostHeight[i],rightMostHeight[i]) - A[i];
			if(curWater > 0){
				water += curWater;
			}
		}
		return water;
    }
};
int main() {
    Solution solution;
    int result;
    int A[] = {0,1,0,2,1,0,1,3,2,1,2,1};
    result = solution.trap(A,12);
    cout<


【代码2】

class Solution {
public:
    int trap(int A[], int n) {
        int *max_left = new int[n]();
        int *max_right = new int[n]();
        for (int i = 1; i < n; i++) {
            max_left[i] = max(max_left[i - 1], A[i - 1]);
            max_right[n - 1 - i] = max(max_right[n - i], A[n - i]);
        }
        int sum = 0;
        for (int i = 0; i < n; i++) {
            int height = min(max_left[i], max_right[i]);
            if (height > A[i]) {
                sum += height - A[i];
            }
        }
        delete[] max_left;
        delete[] max_right;
        return sum;
    }
};


【代码3】

思路2

class Solution {
public:
    //时间复杂度 O(n),空间复杂度 O(1)
    int trap(int A[], int n) {
        // 最高的柱子,将数组分为两半
        int max = 0;
        for (int i = 0; i < n; i++){
            if (A[i] > A[max]) max = i;
        }
        int water = 0;
        for (int i = 0, leftMaxHeight = 0; i < max; i++){
            if (A[i] > leftMaxHeight){
                leftMaxHeight = A[i];
            }
            else {
                water += leftMaxHeight - A[i];
            }
        }
        for (int i = n - 1, rightMaxHeight = 0; i > max; i--){
            if (A[i] > rightMaxHeight){
                rightMaxHeight = A[i];
            }
            else{
                water += rightMaxHeight - A[i];
            }
        }
        return water;
    }
};

【代码4】

//第4种解法,用一个栈辅助,小于栈顶的元素压入,大于等于栈顶就把栈里所有小于或等于当
//前值的元素全部出栈处理掉。
// LeetCode, Trapping Rain Water
// 用一个栈辅助,小于栈顶的元素压入,大于等于栈顶就把栈里所有小于或
// 等于当前值的元素全部出栈处理掉,计算面积,最后把当前元素入栈
// 时间复杂度 O(n),空间复杂度 O(n)
class Solution {
public:
    int trap(int a[], int n) {
        stack> s;
        int water = 0;
        for (int i = 0; i < n; ++i) {
            int height = 0;
            // 将栈里比当前元素矮或等高的元素全部处理掉
            while (!s.empty()) {
                int bar = s.top().first;
                int pos = s.top().second;
                // bar, height, a[i] 三者夹成的凹陷
                water += (min(bar, a[i]) - height) * (i - pos - 1);
                height = bar;
                if (a[i] < bar) // 碰到了比当前元素高的,跳出循环
                    break;
                else
                s.pop(); // 弹出栈顶,因为该元素处理完了,不再需要了
            }
            s.push(make_pair(a[i], i));
        }
        return water;
    }
};


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