54 lines
2.2 KiB
Markdown
54 lines
2.2 KiB
Markdown
---
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id: "2019-01-02-17-15"
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title: "Q807 保持城市天际线(Max Increase to Keep City Skyline)"
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headWord: "leetcode刷题，题目详情见<a href='https://leetcode-cn.com/problems/max-increase-to-keep-city-skyline/'>点击跳转</a>"
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tags: ["java", "leetcode","array",'leetcode']
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categories:
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- "算法"
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- "leetcode刷题"
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---
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### 解析思路
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  leetcode 中等难度中比较简单的一个，题目描述[看这里](https://leetcode-cn.com/problems/max-increase-to-keep-city-skyline/),最开始看题目是比较懵逼的，第一遍没有看懂，多看几次才明白，其实就是让每个点增大到不超过本行且本列的最大值，输出增加的值的和。so 解题方法就是算出每行每列的最大值，然后每个点与当前行当前列最大值中较小的一个差值的和就是结果。
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### 代码(Java实现)
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```java
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class Solution {
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public int maxIncreaseKeepingSkyline(int[][] grid) {
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int rowLength = grid.length;
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int columnLength = grid[0].length;
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//每行对应的最大值
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int[] rowMax = new int[rowLength];
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//每列对应的最大值
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Arrays.fill(rowMax, -1);
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int[] columnMax = new int[columnLength];
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Arrays.fill(columnMax, -1);
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int sum = 0;
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for (int i = 0; i < rowLength; i++) {
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for (int j = 0; j < columnLength; j++) {
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if (rowMax[i] == -1) {
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//计算行最大值
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for (int temp = 0; temp < columnLength; temp++) {
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if (rowMax[i] < grid[i][temp]) {
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rowMax[i] = grid[i][temp];
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}
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}
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}
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if (columnMax[j] == -1) {
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//计算列最大值
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for (int temp = 0; temp < rowLength; temp++) {
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if (columnMax[j] < grid[temp][j]) {
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columnMax[j] = grid[temp][j];
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}
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}
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}
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sum += Math.min(rowMax[i], columnMax[j]) - grid[i][j];
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}
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}
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return sum;
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}
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}
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```
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