Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the sameelement twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
中文版:
給定一個整數數組和一個目標值，找出數組中和爲目標值的兩個數。
你可以假設每個輸入只對應一種答案，且同樣的元素不能被重複利用。
示例:
給定 nums = [2, 7, 11, 15], target = 9
因爲 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]
OC解
//Two Sum
//把數組元素和index生成對應的鍵值對,然後遍歷數組元素若是sum - 元素 字典存在value則 value就是index
NSArray *numArray = @[@(1),@(3),@(5),@(4),@(6)];
NSInteger sum = 12;
NSMutableDictionary *tempDict = [NSMutableDictionary dictionary];
for (int i=0; i<[numArray count]; i++) {
[tempDict setValue:@(i) forKey:[NSString stringWithFormat:@"%ld",[numArray[i] integerValue]]];
}
__block BOOL isHaveValueNumber = NO;
[numArray enumerateObjectsUsingBlock:^(id _Nonnull obj, NSUInteger idx, BOOL * _Nonnull stop) {
NSString *keyStr = [NSString stringWithFormat:@"%ld",sum - [(NSNumber *)obj integerValue]];
NSNumber* lastIndex = tempDict[keyStr];
if (lastIndex && idx < [lastIndex integerValue]) {
NSLog(@"兩個數值的index爲%ld,%ld",[lastIndex integerValue],idx);
isHaveValueNumber = YES;
}
}];
if (!isHaveValueNumber) {
NSLog(@"No vaild outputs");
}