71 lines
1.9 KiB
Java
71 lines
1.9 KiB
Java
//给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。
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//
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//
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//
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// 示例 1：
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//
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//
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//
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//
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//输入：height = [0,1,0,2,1,0,1,3,2,1,2,1]
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//输出：6
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//解释：上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。
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//
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//
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// 示例 2：
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//
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//
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//输入：height = [4,2,0,3,2,5]
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//输出：9
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//
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//
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//
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//
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// 提示：
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//
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//
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// n == height.length
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// 0 <= n <= 3 * 104
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// 0 <= height[i] <= 105
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//
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// Related Topics 栈 数组 双指针 动态规划
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// 👍 2384 👎 0
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package leetcode.editor.cn;
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//42:接雨水
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public class TrappingRainWater {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new TrappingRainWater().new Solution();
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System.out.println(solution.trap(new int[]{0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1}));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int trap(int[] height) {
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int size = height.length;
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if (size < 2) {
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return 0;
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}
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int[] left = new int[size];
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left[0] = height[0];
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for (int i = 1; i < size; i++) {
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left[i] = Math.max(left[i - 1], height[i]);
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}
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int[] right = new int[size];
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right[size - 1] = height[size - 1];
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for (int i = size - 2; i >= 0; i--) {
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right[i] = Math.max(right[i + 1], height[i]);
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}
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int count = 0;
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for (int i = 0; i < size; i++) {
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count += Math.min(left[i], right[i]) - height[i];
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}
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return count;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |