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** 方法一:贪心 + 排序**
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- 生长时间越长的种子,越先播种,因此将 $growTime$ 降序排列。
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+ 根据题目描述,我们知道,每一天只能为一枚种子进行播种,因此不管什么播种顺序,所有种子的播种时间之和总是等于 $\sum_ {i=0}^{n-1} plantTime[ i] $。那么,为了让尽快让所有种子开花,我们应该尽快播种生长时间最长的种子。因此,我们可以对所有种子按照生长时间从大到小进行排序,然后依次进行播种。
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+
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+ 时间复杂度 $O(n \times \log n)$,空间复杂度 $O(n)$。其中 $n$ 是种子的数量。
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<!-- tabs:start -->
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class Solution :
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def earliestFullBloom (self , plantTime : List[int ], growTime : List[int ]) -> int :
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ans = t = 0
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- for a, b in sorted (zip (plantTime, growTime), key = lambda x : - x[1 ]):
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- t += a
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- ans = max (ans, t + b )
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+ for pt, gt in sorted (zip (plantTime, growTime), key = lambda x : - x[1 ]):
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+ t += pt
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+ ans = max (ans, t + gt )
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return ans
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```
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@@ -94,16 +96,15 @@ class Solution:
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class Solution {
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public int earliestFullBloom (int [] plantTime , int [] growTime ) {
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int n = plantTime. length;
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- int [][] arr = new int [n][ 2 ];
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- for (int i = 0 ; i < n; ++ i ) {
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- arr [i] = new int [] {plantTime[i], growTime[i]} ;
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+ Integer [] idx = new Integer [n ];
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+ for (int i = 0 ; i < n; i ++ ) {
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+ idx [i] = i ;
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}
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- Arrays . sort(arr, (a, b) - > b[1 ] - a[1 ]);
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- int ans = 0 ;
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- int t = 0 ;
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- for (int [] e : arr) {
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- t += e[0 ];
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- ans = Math . max(ans, t + e[1 ]);
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+ Arrays . sort(idx, (i, j) - > growTime[j] - growTime[i]);
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+ int ans = 0 , t = 0 ;
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+ for (int i : idx) {
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+ t += plantTime[i];
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+ ans = Math . max(ans, t + growTime[i]);
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}
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return ans;
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}
@@ -117,13 +118,13 @@ class Solution {
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public:
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int earliestFullBloom(vector<int >& plantTime, vector<int >& growTime) {
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int n = plantTime.size();
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- vector<pair< int, int>> arr ;
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- for (int i = 0; i < n; ++i) arr.push_back({-growTime [ i ] , plantTime [ i ] } );
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- sort(arr .begin(), arr .end());
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+ vector<int > idx(n) ;
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+ iota(idx.begin(), idx.end(), 0 );
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+ sort(idx .begin(), idx .end(), [ & ] (int i, int j) { return growTime [ j ] < growTime [ i ] ; } );
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int ans = 0, t = 0;
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- for (auto [ a, b ] : arr ) {
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- t += b ;
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- ans = max(ans, t - a );
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+ for (int i : idx ) {
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+ t += plantTime [ i ] ;
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+ ans = max(ans, t + growTime [ i ] );
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}
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return ans;
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}
@@ -133,20 +134,19 @@ public:
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### **Go**
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```go
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- func earliestFullBloom(plantTime []int, growTime []int) int {
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- arr := [][]int{}
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- for i, a := range plantTime {
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- arr = append(arr, []int{a, growTime[i]})
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+ func earliestFullBloom(plantTime []int, growTime []int) (ans int) {
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+ n := len(plantTime)
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+ idx := make([]int, n)
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+ for i := range idx {
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+ idx[i] = i
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}
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- sort.Slice(arr, func(i, j int) bool {
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- return arr[i][1] > arr[j][1]
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- })
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- ans, t := 0, 0
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- for _, e := range arr {
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- t += e[0]
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- ans = max(ans, t+e[1])
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+ sort.Slice(idx, func(i, j int) bool { return growTime[idx[j]] < growTime[idx[i]] })
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+ t := 0
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+ for _, i := range idx {
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+ t += plantTime[i]
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+ ans = max(ans, t+growTime[i])
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}
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- return ans
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+ return
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}
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func max(a, b int) int {
@@ -159,10 +159,36 @@ func max(a, b int) int {
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### ** TypeScript**
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- <!-- 这里可写当前语言的特殊实现逻辑 -->
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-
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``` ts
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+ function earliestFullBloom(plantTime : number [], growTime : number []): number {
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+ const n = plantTime .length ;
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+ const idx: number [] = Array .from ({ length: n }, (_ , i ) => i );
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+ idx .sort ((i , j ) => growTime [j ] - growTime [i ]);
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+ let [ans, t] = [0 , 0 ];
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+ for (const i of idx ) {
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+ t += plantTime [i ];
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+ ans = Math .max (ans , t + growTime [i ]);
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+ }
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+ return ans ;
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+ }
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+ ```
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+ ### ** Rust**
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+
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+ ``` rust
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+ impl Solution {
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+ pub fn earliest_full_bloom (plant_time : Vec <i32 >, grow_time : Vec <i32 >) -> i32 {
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+ let mut idx : Vec <usize > = (0 .. plant_time . len ()). collect ();
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+ idx . sort_by_key (| & i | - & grow_time [i ]);
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+ let mut ans = 0 ;
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+ let mut t = 0 ;
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+ for & i in & idx {
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+ t += plant_time [i ];
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+ ans = ans . max (t + grow_time [i ]);
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+ }
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+ ans
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+ }
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+ }
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```
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### ** ...**
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