@@ -471,5 +471,73 @@ int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
471471}
472472```
473473
474+ ### Scala
475+ 476+ 暴力解法:
477+ 478+ ``` scala
479+ object Solution {
480+ def canCompleteCircuit (gas : Array [Int ], cost : Array [Int ]): Int = {
481+ for (i <- cost.indices) {
482+ var rest = gas(i) - cost(i)
483+ var index = (i + 1 ) % cost.length // index为i的下一个节点
484+ while (rest > 0 && i != index) {
485+ rest += (gas(index) - cost(index))
486+ index = (index + 1 ) % cost.length
487+ }
488+ if (rest >= 0 && index == i) return i
489+ }
490+ - 1
491+ }
492+ }
493+ ```
494+ 495+ 贪心算法,方法一:
496+ 497+ ``` scala
498+ object Solution {
499+ def canCompleteCircuit (gas : Array [Int ], cost : Array [Int ]): Int = {
500+ var curSum = 0
501+ var min = Int .MaxValue
502+ for (i <- gas.indices) {
503+ var rest = gas(i) - cost(i)
504+ curSum += rest
505+ min = math.min(min, curSum)
506+ }
507+ if (curSum < 0 ) return - 1 // 情况1: gas的总和小于cost的总和,不可能到达终点
508+ if (min >= 0 ) return 0 // 情况2: 最小值>=0,从0号出发可以直接到达
509+ // 情况3: min为负值,从后向前看,能把min填平的节点就是出发节点
510+ for (i <- gas.length - 1 to 0 by - 1 ) {
511+ var rest = gas(i) - cost(i)
512+ min += rest
513+ if (min >= 0 ) return i
514+ }
515+ - 1
516+ }
517+ }
518+ ```
519+ 520+ 贪心算法,方法二:
521+ 522+ ``` scala
523+ object Solution {
524+ def canCompleteCircuit (gas : Array [Int ], cost : Array [Int ]): Int = {
525+ var curSum = 0
526+ var totalSum = 0
527+ var start = 0
528+ for (i <- gas.indices) {
529+ curSum += (gas(i) - cost(i))
530+ totalSum += (gas(i) - cost(i))
531+ if (curSum < 0 ) {
532+ start = i + 1 // 起始位置更新
533+ curSum = 0 // curSum从0开始
534+ }
535+ }
536+ if (totalSum < 0 ) return - 1 // 说明怎么走不可能跑一圈
537+ start
538+ }
539+ }
540+ ```
541+ 474542-----------------------
475543<div align =" center " ><img src =https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width =500 > </img ></div >
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