Trapping Rain Water

標籤：java   leetcode   數組   

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.

方法

對於每一個A[i]，trap water 為 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;    }