LeetCode每日一题(1996. The Number of Weak Characters in the Game)

You are playing a game that contains multiple characters, and each of the characters has two main properties: attack and defense. You are given a 2D integer array properties where properties[i] = [attacki, defensei] represents the properties of the ith character in the game.

A character is said to be weak if any other character has both attack and defense levels strictly greater than this character’s attack and defense levels. More formally, a character i is said to be weak if there exists another character j where attackj > attacki and defensej > defensei.

Return the number of weak characters.

Example 1:

Input: properties = [[5,5],[6,3],[3,6]]
Output: 0

Explanation: No character has strictly greater attack and defense than the other.

Example 2:

Input: properties = [[2,2],[3,3]]
Output: 1
Explanation: The first character is weak because the second character has a strictly greater attack and defense.

Example 3:

Input: properties = [[1,5],[10,4],[4,3]]
Output: 1

Explanation: The third character is weak because the second character has a strictly greater attack and defense.

Constraints:

• 2 <= properties.length <= 105
• properties[i].length == 2
• 1 <= attacki, defensei <= 105

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1. 根据 attack 排序， 如果 attack 相同则按 defense 的倒序排序
2. 反向遍历，时刻记录遍历过的最大的 defense 值， 如果 properties[i][1] < max(defense)则认为 properties[i]弱于前面便利过的某个 property

之所以 attack 相同时按 defense 的倒序排序， 是因为题目要求是 attack 和 defense 都严格小于才能算弱， 我们这样排序可以保证 attack 是不严格递增， 按 defense 倒序排序可以保证在遍历的时候可以避免误把 attack 相同的情况计入答案中

---

impl Solution {
    pub fn number_of_weak_characters(mut properties: Vec<Vec<i32>>) -> i32 {
        properties.sort_by(|v1, v2| {
            if v1[0] == v2[0] {
                return v2[1].cmp(&v1[1]);
            }
            v1[0].cmp(&v2[0])
        });
        let mut max = properties.last().unwrap()[1];
        let mut ans = 0;
        for v in properties.into_iter().rev().skip(1) {
            if v[1] < max {
                ans += 1;
            }
            max = max.max(v[1]);
        }
        ans
    }
}
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