[LeetCode]保持城市天际线

这个问题思路很简单主要是提升效率,把水平天际线和垂直的天际线放到一个循环里面去求值,这样可以省下一个O(n*n)的时间,然后求增加总量的时候,同样只用一个循环。

问题链接:https://leetcode-cn.com/problems/max-increase-to-keep-city-skyline/

代码:

    public int maxIncreaseKeepingSkyline(int[][] grid) {
        int result = 0;
        int[] vertical = new int[grid[0].length];  //垂直
        int[] horizontal = new int[grid.length];  //水平
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                vertical[j] = Math.max(grid[i][j],vertical[j]);
                horizontal[i] = Math.max(grid[i][j],horizontal[i]);
            }
        }
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                result+=Math.min(horizontal[i],vertical[j])-grid[i][j];
            }
        }
        return result;
    }

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转载自blog.csdn.net/sinat_37273780/article/details/84953443
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