trapping-rain-water

题目：

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], 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 Marcos for contributing this image!

程序：

方法一：

夹逼遍历

class Solution {public:    int trap(int A[], int n) {        if(n <= 2) return 0;        int max = -1, maxInd = 0;        int i = 0;        for(; i < n; ++i){            if(A[i] > max){                max = A[i];                maxInd = i;            }        }        int area = 0, root = A[0];        for(i = 0; i < maxInd; ++i){            if(root < A[i]) root = A[i];            else area += (root - A[i]);        }        for(i = n-1, root = A[n-1]; i > maxInd; --i){            if(root < A[i]) root = A[i];            else area += (root - A[i]);        }        return area;    }};

方法二：

左右遍历数组

public int trap(int[] A) {    if(A==null || A.length==0)        return 0;    int max = 0;    int res = 0;    int[] container = new int[A.length];    for(int i=0;i<A.length;i++)    {        container[i]=max;        max = Math.max(max,A[i]);    }    max = 0;    for(int i=A.length-1;i>=0;i--)    {        container[i] = Math.min(max,container[i]);        max = Math.max(max,A[i]);        res += container[i]-A[i]>0?container[i]-A[i]:0;    }        return res;}

点评：

在做数组类题的时候夹逼和左右遍历数组都是常用方法