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  • [leetcode] Trapping Rain Water

    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!

    https://oj.leetcode.com/problems/trapping-rain-water/

    思 路1:对某个值A[i]来说,能装的最多的水取决于在i之前最高的值leftMaxt[i]和在i右边的最高的rightMax[i],容水量即 min(left,right) – A[i]。:为了计算高度,第一遍从左到右计算数组leftMaxt,第二遍从右到左计算rightMax。时空复杂度都是O(n)。

    思路2:改进思路1,可以用常量空间搞定。具体思路,先找到最高的bar,然后总两边向中间计算,边计算边更新curHeight。

    具体可参考 这里

    思路1:

    public class Solution {
    	public int trap(int[] A) {
    		if (A == null || A.length < 3)
    			return 0;
    		int n = A.length;
    		int[] leftMax = new int[n];
    		int[] rightMax = new int[n];
    
    		int i;
    		int max = A[0];
    		for (i = 1; i < n - 1; i++) {
    			leftMax[i] = max;
    			if (A[i] > max)
    				max = A[i];
    
    		}
    
    		max = A[n - 1];
    		for (i = n - 2; i > 0; i--) {
    			rightMax[i] = max;
    			if (A[i] > max)
    				max = A[i];
    
    		}
    
    		int count = 0;
    
    		for (i = 1; i < n - 1; i++) {
    			int min = leftMax[i] < rightMax[i] ? leftMax[i] : rightMax[i];
    			if (min > A[i])
    				count += min - A[i];
    
    		}
    		return count;
    	}
    
    	public static void main(String[] args) {
    		System.out.println(new Solution().trap(new int[] { 0, 1, 0, 2, 1, 0, 1,
    				3, 2, 1, 2, 1 }));
    
    	}
    
    }

    第三遍

    思路二:

    public class Solution {
        public int trap(int[] A) {
            if (A == null || A.length < 3)
                return 0;
            int n = A.length;
            int count=0;
            int midIdx = 0;
            int highest=A[0];
            for(int i=0;i<n;i++){
                if(A[i]>highest){
                    midIdx=i;
                    highest=A[i];
                }   
            }
            
            int curHighest=0;
            for(int i=0;i<midIdx;i++){
                if(A[i]<curHighest){
                    count+=curHighest-A[i];
                }else{
                    curHighest=A[i];
                }
            }
            
            curHighest=0;
            for(int i=n-1;i>midIdx;i--){
                if(A[i]<curHighest){
                    count+=curHighest-A[i];
                }else{
                    curHighest=A[i];
                }    
            }
            
            return count;
            
        }
    }

     

    参考:

    http://fisherlei.blogspot.com/2013/01/leetcode-trapping-rain-water.html

    http://www.cnblogs.com/lichen782/p/Leetcode_Trapping_Rain_Water.html

     

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  • 原文地址:https://www.cnblogs.com/jdflyfly/p/3810751.html
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