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Trapping Rain Water

时间:2014-06-27 23:26:26      阅读:494      评论:0      收藏:0      [点我收藏+]

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], return 6.

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方法

对于每一个A[i],trap water 为 min(left, right)- A[i];
    public int trap(int[] A) {
        if (A == null || A.length == 0 || A.length == 1 || A.length == 2) {
            return 0;
        }
        int len = A.length;
        int[] left = new int[len];
        int[] right = new int[len];
        int leftMax = 0;
        for (int i = 0; i < len; i++) {
        	left[i] = leftMax;
        	if (A[i] > leftMax) {
        		leftMax = A[i];
        	}
        }
        int rightMax = 0;
        for (int i = len; i > 0; i--) {
        	right[i - 1] = rightMax;
        	if (A[i - 1] > rightMax) {
        		rightMax = A[i - 1];
        	}
        }
        
        int sum = 0;
        for (int i = 0; i < len; i++) {
        	int temp = Math.min(left[i], right[i]) - A[i];
        	if (temp > 0) {
        		sum += temp;
        	}
        }
        return sum;
    }


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Trapping Rain Water

原文:http://blog.csdn.net/u010378705/article/details/35314741

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